{"id":754,"date":"2023-04-30T10:00:12","date_gmt":"2023-04-30T04:30:12","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=754"},"modified":"2023-05-01T09:30:12","modified_gmt":"2023-05-01T04:00:12","slug":"distance-between-two-points","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/distance-between-two-points\/","title":{"rendered":"Distance Between Two Points"},"content":{"rendered":"<h2>Distance Between Two Points<\/h2>\n<h3>Distance Between Two Points Formula<\/h3>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127630\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-1.jpg\" alt=\"Distance Between Two Points 1\" width=\"886\" height=\"607\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-1.jpg 886w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-1-300x206.jpg 300w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-1-768x526.jpg 768w\" sizes=\"(max-width: 886px) 100vw, 886px\" \/><\/p>\n<p><strong>(1) Distance between two points on X-axis :<\/strong><br \/>\nThe coordinate axes in the coordinate plane can be treated as number lines.<br \/>\nIf P(x<sub>1<\/sub>, 0) and Q(x<sub>2<\/sub>, 0) are two points on X-axis, the distance between them is taken as<br \/>\nPQ = |x<sub>1<\/sub>-x<sub>2<\/sub>| &#8230;&#8230;&#8230;.(i)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127631\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-2.jpg\" alt=\"Distance Between Two Points 2\" width=\"294\" height=\"307\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-2.jpg 294w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-2-287x300.jpg 287w\" sizes=\"(max-width: 294px) 100vw, 294px\" \/><br \/>\n(2) Distance between two points on Y-axis:<br \/>\nIf the points A(0, y<sub>1<\/sub>) and B(0, y<sub>2<\/sub>) are two points on Y-axis, the distance between them is taken as<br \/>\nAB = |y<sub>1<\/sub>-y<sub>2<\/sub>| &#8230;&#8230;&#8230;.(ii)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127632\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-3.jpg\" alt=\"Distance Between Two Points 3\" width=\"297\" height=\"306\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-3.jpg 297w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-3-291x300.jpg 291w\" sizes=\"(max-width: 297px) 100vw, 297px\" \/><\/p>\n<p><strong>(3) Distance between P(x<sub>1<\/sub>, y<sub>1<\/sub>) and Q(x<sub>2<\/sub>, y<sub>2<\/sub>):<\/strong><br \/>\nLet P(x<sub>1<\/sub>, y<sub>1<\/sub>) and Q(x<sub>2<\/sub>, y<sub>2<\/sub>) be two given points in the coordinate plane.<br \/>\nLet M and N be the feet of perpendiculars from P(x<sub>1<\/sub>, y<sub>1<\/sub>) and Q(x<sub>2<\/sub>, y<sub>2<\/sub>) respectively to X-axis.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127634\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-4.jpg\" alt=\"Distance Between Two Points 4 \" width=\"415\" height=\"345\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-4.jpg 415w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-4-300x249.jpg 300w\" sizes=\"(max-width: 415px) 100vw, 415px\" \/><br \/>\n\u2234 M and N are respectively (x<sub>1<\/sub>, 0) and (x<sub>2<\/sub>, 0).<br \/>\nMN = |x<sub>1<\/sub>-x<sub>2<\/sub>| &#8230;&#8230;&#8230;.(i)<br \/>\nLet R and S be the feet of perpendiculars from P(x<sub>1<\/sub>, y<sub>1<\/sub>) and Q(x<sub>2<\/sub>, y<sub>2<\/sub>) to Y-axis.<br \/>\nR and S are respectively (0, y<sub>1<\/sub>) and (0, y<sub>2<\/sub>).<br \/>\nRS = |y<sub>1<\/sub>-y<sub>2<\/sub>| &#8230;&#8230;&#8230;.(ii)<br \/>\nLet PR and QN intersect in T.<br \/>\nClearly in \u2206PQT, \u2220PTQ is a right angle.<br \/>\nUsing Pythagoras&#8217; theorem we have,<br \/>\nPQ<sup>2<\/sup> = PT<sup>2<\/sup> + QT<sup>2<\/sup> = MN<sup>2<\/sup> + RS<sup>2<\/sup><br \/>\nbecause QTRS and PTNM are rectangles.<br \/>\nNow, by (i) and (ii), we have<br \/>\nPQ<sup>2<\/sup> = MN<sup>2<\/sup> + RS<sup>2<\/sup><br \/>\n=|x<sub>1<\/sub>-x<sub>2<\/sub>|<sup>2<\/sup> + |y<sub>1<\/sub>-y<sub>2<\/sub>|<sup>2<\/sup><br \/>\n= (x<sub>1<\/sub>-x<sub>2<\/sub>)<sup>2<\/sup> + (y<sub>1<\/sub>-y<sub>2<\/sub>)<sup>2<\/sup><br \/>\n\u2234 \\( PQ=\\sqrt{{{\\left( {{x}_{1}}-{{x}_{2}} \\right)}^{2}}+{{\\left( {{y}_{1}}-{{y}_{2}} \\right)}^{2}}} \\)<br \/>\nFormula (iii) gives the distance between two points whose coordinates are (x<sub>1<\/sub>, y<sub>1<\/sub>) and (x<sub>2<\/sub>, y<sub>2<\/sub>). The distance between the points P and Q is also denoted by d(P, Q).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127635\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-45.jpg\" alt=\"Distance Between Two Points 45\" width=\"280\" height=\"298\" \/><br \/>\nThus, d(P, Q) = PQ<br \/>\n= d(P(x<sub>1<\/sub>, y<sub>1<\/sub>), Q(x<sub>2<\/sub>, y<sub>2<\/sub>)) = \\(\\sqrt{{{\\left( {{x}_{1}}-{{x}_{2}} \\right)}^{2}}+{{\\left( {{y}_{1}}-{{y}_{2}} \\right)}^{2}}} \\)<br \/>\nIf P and Q lie on X-axis then also formula remains same.<br \/>\nHere, M = P and N = O.<br \/>\nR = O and S = Q.<br \/>\nMN = OP = |x<sub>1<\/sub>-0| = |x<sub>1<\/sub>| = |x<sub>1<\/sub>-y<sub>1<\/sub>| (y<sub>1<\/sub> = 0)<br \/>\nRS = OQ = |0-y<sub>2<\/sub>| = |x<sub>2<\/sub>-y<sub>2<\/sub>| (x<sub>2<\/sub> = 0)<br \/>\n\u2234 \\(PQ=\\sqrt { { \\left| { { x }_{ 1 } }-{ { x }_{ 2 } } \\right| \u00a0}^{ 2 }+{ \\left| { { y }_{ 1 } }-{ { y }_{ 2 } } \\right| \u00a0}^{ 2 } } =\\sqrt { { { \\left( { { x }_{ 1 } }-{ { x }_{ 2 } } \\right) \u00a0}^{ 2 } }+{ { \\left( { { y }_{ 1 } }-{ { y }_{ 2 } } \\right) \u00a0}^{ 2 } } }\\)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127636\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-6.jpg\" alt=\"Distance Between Two Points 6\" width=\"292\" height=\"303\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-6.jpg 292w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/01\/Distance-Between-Two-Points-6-289x300.jpg 289w\" sizes=\"(max-width: 292px) 100vw, 292px\" \/><br \/>\n[Note : P may lie on Y-axis<br \/>\nHere x<sub>1<\/sub>\u00a0= 0.<br \/>\nHere also PS = |y<sub>1<\/sub>-y<sub>2<\/sub>| (P = R)<br \/>\nMN = = |0-x<sub>2<\/sub>| = |x<sub>2<\/sub>-y<sub>2<\/sub>|<br \/>\nSimilarly if P lies on X-axis, the formula remains same.]<br \/>\nIf PQ is parallel to any axis then x<sub>1<\/sub> = x<sub>2<\/sub> or y<sub>1<\/sub> = y<sub>2<\/sub> and formula remains same.<\/p>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/0IOEPcAHgi4?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p>\u2018Distance\u2019 is also known as a \u2018Metric\u2019. Metric plays an important role in all kinds of geometry, including Euclidean geometry. In fact the nature of Metric defines the type of geometry. The following properties of Metric are not only interesting but also very useful.<br \/>\n(1) d(A, B) = AB \u2265 0 i.e. the distance between two points is a non-negative real number.<br \/>\n(2) d(A, B) = AB = 0, if and only if A = B.<br \/>\n(3) d(A, B) = d(B, A) i.e. AB = BA.<br \/>\n(4) If A(x<sub>1<\/sub>, y<sub>1<\/sub>), B(x<sub>2<\/sub>, y<sub>2<\/sub>), C(x<sub>3<\/sub>, y<sub>3<\/sub>) are three points in coordinate plane, then<br \/>\nd(A, B) + d(B, C) \u2265 d(A, C) i.e. AB + BC \u2265 AC.<br \/>\nIf, A, B, C are collinear points and A\u2014B\u2014C then, AB + BC = AC.<br \/>\nIf, A, B, C are non-collinear points or collinear but B\u2014A\u2014C or A\u2014C\u2014B, then AB + BC &gt; AC<br \/>\nThe formula for PQ can also be written as PQ<sup>2<\/sup> = (x<sub>1<\/sub>-x<sub>2<\/sub>)<sup>2<\/sup> + (y<sub>1<\/sub>-y<sub>2<\/sub>)<sup>2<\/sup>. While solving the examples, this form of distance formula is advantageous. We have to be careful that at the end when we find the distance \u2018PQ\u2019 we have to take the positive square root of the expression for PQ<sup>2<\/sup>.<\/p>\n<p>Read more: <a href=\"https:\/\/www.aplustopper.com\/section-formula\/\">Section Formula<\/a><\/p>\n<h2><strong>Distance Between Two Points With Examples<\/strong><\/h2>\n<p><strong>Example 1:<\/strong> \u00a0 \u00a0Find the distance between two points<br \/>\n(i) P(\u20136, 7) and Q(\u20131, \u20135)<br \/>\n(ii) R(a + b, a \u2013 b) and S(a \u2013 b, \u2013a \u2013 b)<br \/>\n\\(\\left( \\text{iii} \\right)\\text{ A(at}_{1}^{2}\\text{, 2a}{{\\text{t}}_{1}}\\text{) and B(at}_{2}^{2}\\text{, 2a}{{\\text{t}}_{2}}\\text{)}\\)<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0<strong>(i)<\/strong> Here, \u00a0x<sub>1<\/sub>\u00a0= \u2013 6, y<sub>1<\/sub> = 7 and x<sub>2<\/sub> = \u2013 1, y<sub>2<\/sub> = \u2013 5<br \/>\n\u2234 \u00a0\\( PQ=\\sqrt{{{\\left( {{x}_{2}}-{{x}_{1}} \\right)}^{2}}+{{\\left( {{y}_{2}}-{{y}_{1}} \\right)}^{2}}} \\)<br \/>\n\\( \\Rightarrow PQ=\\sqrt{{{(-1+6)}^{2}}+{{(-5-7)}^{2}}} \\\\ \\)<br \/>\n\\( \\Rightarrow PQ=\\sqrt{25+144}=\\sqrt{169}=13 \\)<br \/>\n<strong>(ii)<\/strong> We have,<br \/>\n\\( RS=\\sqrt{{{(a-b-a-b)}^{2}}+{{(-a-b-a+b)}^{2}}} \\)<br \/>\n\\( \\Rightarrow RS=\\sqrt{4{{b}^{2}}+4{{a}^{2}}}=2\\sqrt{{{a}^{2}}+{{b}^{2}}} \\)<br \/>\n<strong>(iii)<\/strong> We have,<br \/>\n\\( AB=\\sqrt{{{(a{{t}_{2}}^{2}-a{{t}_{1}}^{2})}^{2}}+{{(2a{{t}_{2}}-2a{{t}_{1}})}^{2}}} \\)<br \/>\n\\( \\Rightarrow AB=\\sqrt{{{a}^{2}}{{({{t}_{2}}-{{t}_{1}})}^{2}}{{({{t}_{2}}+{{t}_{1}})}^{2}}+4{{a}^{2}}{{({{t}_{2}}-{{t}_{1}})}^{2}}} \\)<br \/>\n\\(\\Rightarrow AB=a({{t}_{2}}-{{t}_{1}})\\sqrt{{{({{t}_{2}}+{{t}_{1}})}^{2}}+4}\\)<\/p>\n<p><strong>Example 2:<\/strong> \u00a0 \u00a0If the point (x, y) is equidistant from the points (a + b, b \u2013 a) and (a \u2013 b, a + b), prove that bx = ay.<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0Let P(x, y), Q(a + b, b \u2013 a) and R (a \u2013 b, a + b) be the given points. Then<br \/>\nPQ = PR\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(Given)<br \/>\n\\(\\Rightarrow \\sqrt{{{\\{x-(a+b)\\}}^{2}}+{{\\{y-(b-a)\\}}^{2}}}=\\sqrt{{{\\{x-(a-b)\\}}^{2}}+{{\\{y-(a+b)\\}}^{2}}}\\)<br \/>\n\u21d2 {x \u2013 (a + b)}<sup>2<\/sup> + {y \u2013 (b \u2013 a)}<sup>2<\/sup> = {x \u2013 (a \u2013 b)}<sup>2<\/sup> + {y \u2013 (a + b)}<sup>2<\/sup><br \/>\n\u21d2 x<sup>2<\/sup>\u00a0\u2013 2x (a + b) + (a + b)<sup>2<\/sup> + y<sup>2<\/sup> \u2013 2y (b \u2013 a) + (b \u2013 a)<sup>2<\/sup> = x<sup>2<\/sup>\u00a0+ (a \u2013 b)<sup>2<\/sup> \u2013 2x(a \u2013 b) + y<sup>2<\/sup> \u2013 2 (a + b) + (a + b)<sup>2<\/sup><br \/>\n\u21d2 \u20132x (a + b) \u2013 2y (b \u2013 a) = \u2013 2x (a \u2013 b) \u2013 2y (a + b)<br \/>\n\u21d2 ax + bx + by \u2013 ay = ax \u2013 bx + ay + by<br \/>\n\u21d2 2bx = 2ay \u21d2\u00a0bx = ay<\/p>\n<p><strong>Example 3:<\/strong> \u00a0 \u00a0Find the value of x, if the distance between the points (x, \u2013 1) and (3, 2) is 5.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Let P(x, \u2013 1) and Q(3, 2) be the given points, Then,<br \/>\nPQ = 5 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Given)<br \/>\n\\(\\Rightarrow \\sqrt{{{(x-3)}^{2}}+{{(-1-2)}^{2}}}=5\\)<br \/>\n\u21d2 (x \u2013 3)<sup>2<\/sup> + 9 = 5<sup>2<\/sup><br \/>\n\u21d2 x<sup>2<\/sup>\u00a0\u2013 6x + 18 = 25 \u00a0\u00a0\u21d2 \u00a0x<sup>2<\/sup>\u00a0\u2013 6x \u2013 7 = 0<br \/>\n\u21d2 (x \u2013 7) (x + 1) = 0 \u00a0\u00a0\u21d2 \u00a0x = 7 or x = \u2013 1<\/p>\n<p><strong>Example 4:<\/strong>\u00a0 \u00a0 \u00a0Show that the points (a, a), (\u2013a, \u2013a) and (\u2013\u00a0\u221a3 a, \u221a3 a) are the vertices of an equilateral triangle. Also find its area.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Let A (a, a), B(\u2013a, \u2013a) and C(\u2013\u00a0\u221a3 a, \u221a3 a) be the given points. Then, we have<br \/>\n\\( AB=\\sqrt{{{(-a-a)}^{2}}+{{(-a-a)}^{2}}}=\\sqrt{4{{a}^{2}}+4{{a}^{2}}}=2\\sqrt{2}a \\)<br \/>\n\\( BC=\\sqrt{{{(-\\sqrt{3}a+a)}^{2}}+{{(\\sqrt{3}a+a)}^{2}}} \\)<br \/>\n\\( \\Rightarrow BC=\\sqrt{{{a}^{2}}{{(1-\\sqrt{3})}^{2}}+{{a}^{2}}{{(\\sqrt{3}+1)}^{2}}} \\)<br \/>\n\\( \\Rightarrow BC=a\\sqrt{1+3-2\\sqrt{3}+1+3+2\\sqrt{3}}=a\\sqrt{8}=2\\sqrt{2}a \\)<br \/>\n\\( and,\\text{ }AC=\\sqrt{(-\\sqrt{3}a-{{a}^{2}})+{{(\\sqrt{3}a-a)}^{2}}} \\\\ \\)<br \/>\n\\( \\Rightarrow AC=\\sqrt{{{a}^{2}}(\\sqrt{3}+1)+{{a}^{2}}{{(\\sqrt{3}-1)}^{2}}} \\)<br \/>\n\\( \\Rightarrow AC=\\sqrt{3+1+2\\sqrt{3}+3+1-2\\sqrt{3}}=a\\sqrt{8}=2\\sqrt{2}a \\)<br \/>\nClearly, we have<br \/>\nAB = BC = AC<br \/>\nHence, the triangle ABC formed by the given points is an equilateral triangle.<br \/>\nNow,<br \/>\nArea of \u2206ABC = \\(\\frac { \\sqrt { 3 } \u00a0}{ 4 } \\) \u00a0(side)<sup>2<\/sup><br \/>\n\u21d2 Area of \u2206ABC = \\(\\frac { \\sqrt { 3 } \u00a0}{ 4 } \\) \u00d7 AB<sup>2<\/sup><br \/>\n\u21d2 Area of \u2206ABC = \\(\\frac { \\sqrt { 3 } \u00a0}{ 4 } \\) \u00d7 (2\u221a2 a)<sup>2<\/sup> sq. units = 2\u221a3 a<sup>2<\/sup> sq. units<\/p>\n<p><strong>Example 5:<\/strong>\u00a0 \u00a0 \u00a0Show that the points (1, \u2013 1), (5, 2) and (9, 5) are collinear.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let A (1, \u20131), B (5, 2) and C (9, 5) be the given points. Then, we have\u00a0Clearly, AC = AB + BC<br \/>\n\\( AB=\\sqrt{{{(5-1)}^{2}}+{{(2+1)}^{2}}}=\\sqrt{16+9}=5 \\)<br \/>\n\\( BC=\\sqrt{{{(5-9)}^{2}}+{{(2-5)}^{2}}}=\\sqrt{16+9}=5 \\)<br \/>\n\\( and,~~~~~AC=\\sqrt{{{(1-9)}^{2}}+{{(-1-5)}^{2}}}=\\sqrt{64+36}=10 \\)<br \/>\nHence, A, B, C are collinear points.<\/p>\n<p><strong>Example 6:<\/strong>\u00a0 \u00a0 \u00a0Show that four points (0, \u2013 1), (6, 7), (\u20132, 3) and (8, 3) are the vertices of a rectangle. Also, find its area.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Let A (0, \u20131), B(6, 7), C(\u20132, 3) and D (8, 3) be the given points. Then,<br \/>\n\\( AD=\\sqrt{{{(8-0)}^{2}}+{{(3+1)}^{2}}}=\\sqrt{64+16}=4\\sqrt{5} \\)<br \/>\n\\( BC=\\sqrt{{{(6+2)}^{2}}+{{(7-3)}^{2}}}=\\sqrt{64+16}=4\\sqrt{5} \\)<br \/>\n\\( AC=\\sqrt{{{(-2-0)}^{2}}+{{(3+1)}^{2}}}=\\sqrt{4+16}=2\\sqrt{5} \\)<br \/>\n\\( and,\\text{ }BD\\text{ }=\\sqrt{{{(8-6)}^{2}}+{{(3-7)}^{2}}}=\\sqrt{4+16}=2\\sqrt{5} \\)<br \/>\n\u2234 AD = BC and AC = BD.<br \/>\nSo, ADBC is a parallelogram,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127644\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-7.png\" alt=\"Distance Between Two Points 7\" width=\"323\" height=\"130\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-7.png 323w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-7-300x121.png 300w\" sizes=\"(max-width: 323px) 100vw, 323px\" \/><br \/>\n\\( Now,\\text{ }AB\\text{ }=\\sqrt{{{(6-0)}^{2}}+{{(7+1)}^{2}}}=\\sqrt{36+64}=10 \\)<br \/>\n\\( and~~~~~~CD\\text{ }=\\sqrt{{{(8+2)}^{2}}+{{(3-3)}^{2}}}=10 \\)<br \/>\nClearly, AB<sup>2<\/sup> = AD<sup>2<\/sup> + DB<sup>2<\/sup> and CD<sup>2<\/sup> = CB<sup>2<\/sup> + BD<sup>2<\/sup><br \/>\nHence, ADBC is a rectangle.<br \/>\nNow, Area of rectangle ADBC = AD \u00d7 DB<br \/>\n= (4\u221a5 \u00d7 2\u221a5 ) sq. units = 40 sq. units<\/p>\n<p><strong>Example 7:<\/strong>\u00a0 \u00a0 \u00a0If P and Q are two points whose coordinates are (at<sup>2<\/sup>, 2at) and (a\/t<sup>2<\/sup>, 2a\/t) respectively and S is the point (a, 0). Show that \u00a0 \u00a0\\(\\frac{1}{SP}+\\frac{1}{SQ}\\) \u00a0is independent of t.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> We have,<br \/>\n\\( SP=\\sqrt{{{(a{{t}^{2}}-a)}^{2}}+{{(2at-0)}^{2}}} \\)<br \/>\n\\( =\\sqrt{{{({{t}^{2}}-1)}^{2}}+4{{t}^{2}}}=a({{t}^{2}}+\\text{ }1) \\)<br \/>\n\\( and~~~~~~SQ\\text{ }=\\sqrt{{{\\left( \\frac{a}{{{t}^{2}}}-a \\right)}^{2}}+{{\\left( \\frac{2a}{t}-0 \\right)}^{2}}} \\)<br \/>\n\\( \\Rightarrow SQ\\text{ }=\\sqrt{\\frac{{{a}^{2}}{{(1-{{t}^{2}})}^{2}}}{{{t}^{4}}}+\\frac{4{{a}^{2}}}{{{t}^{2}}}} \\)<br \/>\n\\( \\Rightarrow SQ\\text{ }=\\frac{a}{{{t}^{2}}}\\sqrt{{{(1-{{t}^{2}})}^{2}}+4{{t}^{2}}}=\\frac{a}{{{t}^{2}}}\\sqrt{{{(1+{{t}^{2}})}^{2}}} \\)<br \/>\nwhich is independent of t.<br \/>\n\\( =~\\frac{a}{{{t}^{2}}}(1\\text{ }+\\text{ }{{t}^{2}})~ \\)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127667\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-14.png\" alt=\"Distance Between Two Points 14\" width=\"333\" height=\"72\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-14.png 333w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-14-300x65.png 300w\" sizes=\"(max-width: 333px) 100vw, 333px\" \/><br \/>\n\\( \\Rightarrow \\frac{1}{SP}+\\frac{1}{SQ}=\\frac{1+{{t}^{2}}}{a({{t}^{2}}+1)}=\\frac{1}{a} \\)<\/p>\n<p><strong>Example 8:<\/strong>\u00a0 \u00a0 \u00a0If two vertices of an equilateral triangle be (0, 0), (3, \u221a3 ), find the third vertex.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> O(0, 0) and A(3, \u221a3) be the given points and let B(x, y) be the third vertex of equilateral \u2206OAB. Then, OA = OB = AB<br \/>\n\u21d2 \u00a0OA<sup>2<\/sup> = OB<sup>2<\/sup> = AB<sup>2<\/sup><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127660\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-8.png\" alt=\"Distance Between Two Points 8\" width=\"254\" height=\"198\" \/><br \/>\nWe have, OA<sup>2\u00a0<\/sup>= (3 \u2013 0)<sup>2<\/sup> + (\u221a3 \u2013 0)<sup>2<\/sup> = 12,<br \/>\nOB<sup>2\u00a0<\/sup>= x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup><br \/>\nand, AB<sup>2\u00a0<\/sup>= (x \u2013 3)<sup>2<\/sup>+ (y \u2013 \u221a3)<sup>2<\/sup><br \/>\n\u21d2 \u00a0AB<sup>2<\/sup> = x<sup>2<\/sup> + y<sup>2<\/sup> \u2013 6x \u2013 2 y + 12<br \/>\n\u2234 OA<sup>2<\/sup> = OB<sup>2<\/sup> = AB<sup>2<\/sup><br \/>\n\u21d2 \u00a0OA<sup>2<\/sup> = OB<sup>2<\/sup> and OB<sup>2<\/sup> = AB<sup>2<\/sup><br \/>\n\u21d2 \u00a0x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup> = 12<br \/>\nand, x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup> = x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup> \u2013 6x \u2013 2 \u221a3y + 12<br \/>\n\u21d2 \u00a0x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup> = 12 and 6x + 2 \u221a3y = 12<br \/>\n\u21d2 \u00a0x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup> = 12 and 3x + \u221a3y = 6<br \/>\n\\( \\Rightarrow \\text{ }~{{x}^{2}}~+\\text{ }{{\\left( \\frac{6-3x}{\\sqrt{3}} \\right)}^{2}}=\\text{ }12\\text{ }\\left[ \\because \\ 3x+\\sqrt{3}y=6\\ \\ \\therefore \\ \\ y=\\frac{6-3x}{\\sqrt{3}} \\right] \\)<br \/>\n\u21d2 \u00a03x<sup>2<\/sup> + (6 \u2013 3x)<sup>2<\/sup> = 36<br \/>\n\u21d2 \u00a012x<sup>2<\/sup> \u2013 36x = 0 \u00a0 \u00a0 \u21d2 \u00a0 \u00a0 x = 0, 3<br \/>\n\u2234 \u00a0x = 0 \u00a0\u21d2 \u00a0\u00a0\u221a3y = 6<br \/>\n\\( \\Rightarrow \\text{ }y=\\frac{6}{\\sqrt{3}}=2\\sqrt{3}\\text{ }\\left[ \\text{Putting}\\ \\text{x}=\\text{0}\\ \\text{in 3x + }\\sqrt{\\text{3}}\\text{y}=\\text{6} \\right] \\)<br \/>\nand, x = 3 \u21d2 \u00a0\u00a09 +\u00a0\u221a3 y = 6<br \/>\n\\(\\Rightarrow \\text{ }y=\\frac{6-9}{\\sqrt{3}}=-\\sqrt{3}\\text{ \u00a0 \u00a0}\\left[ \\text{Putting}\\ \\text{x}=\\text{0}\\ \\text{in 3x + }\\sqrt{\\text{3}}\\text{y}=\\text{6} \\right]\\)<br \/>\nHence, the coordinates of the third vertex B are (0, 2 \u221a3) or (3, \u2013 \u221a3).<\/p>\n<p><strong>Example 9:<\/strong>\u00a0 \u00a0 \u00a0Find the coordinates of the circumcentre of the triangle whose vertices are (8, 6), (8, \u2013 2) and (2, \u2013 2). Also, find its circum radius.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Recall that the circumcentre of a triangle is equidistant from the vertices of a triangle. Let A (8, 6), B(8, \u20132) and C(2, \u2013 2) be the vertices of the given triangle and let P (x, y) be the circumcentre of this triangle. Then,<br \/>\nPA = PB = PC \u21d2 \u00a0\u00a0PA<sup>2<\/sup> = PB<sup>2<\/sup> = PC<sup>2<\/sup><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127661\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-9.png\" alt=\"Distance Between Two Points 9\" width=\"337\" height=\"261\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-9.png 337w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-9-300x232.png 300w\" sizes=\"(max-width: 337px) 100vw, 337px\" \/><br \/>\nNow, PA<sup>2<\/sup> = PB<sup>2<\/sup><br \/>\n\u21d2 \u00a0(x \u2013 8)<sup>2<\/sup> + (y \u2013 6)<sup>2<\/sup> = (x \u2013 8)2 + (y + 2)<sup>2<\/sup><br \/>\n\u21d2 \u00a0x<sup>2<\/sup> + y<sup>2<\/sup> \u2013 16x \u2013 12y + 100 = x<sup>2<\/sup> + y<sup>2<\/sup> \u2013 16x + 4y + 68<br \/>\n\u21d2 \u00a016y = 32 \u21d2 \u00a0\u00a0y = 2<br \/>\nand, PB<sup>2<\/sup> = PC<sup>2<\/sup><br \/>\n\u21d2 \u00a0(x \u2013 8)<sup>2<\/sup> + (y + 2)<sup>2<\/sup> = (x \u2013 2)<sup>2<\/sup> + (y + 2)<sup>2<\/sup><br \/>\n\u21d2 \u00a0x<sup>2<\/sup> + y<sup>2<\/sup> \u2013 16x + 4y + 68 = x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup> \u2013 4x + 4y + 8<br \/>\n\u21d2 \u00a012x = 60 \u21d2 \u00a0\u00a0x = 5<br \/>\nSo, the coordinates of the circumcentre P are (5, 2).<br \/>\nAlso, Circum-radius = PA = PB = PC<br \/>\n\\(=\\sqrt{{{(5-8)}^{2}}+{{(2-6)}^{2}}}=\\sqrt{9+16}=5\\)<\/p>\n<p><strong>Example 10:<\/strong>\u00a0 \u00a0 \u00a0If the opposite vertices of a square are (1, \u2013 1) and (3, 4), find the coordinates of the remaining angular points.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let A(1, \u2013 1) and C(3, 4) be the two opposite vertices of a square ABCD and let B(x, y) be the third vertex.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127662\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-10.png\" alt=\"Distance Between Two Points 10\" width=\"219\" height=\"154\" \/><br \/>\nThen, AB = BC<br \/>\n\u21d2 \u00a0AB<sup>2<\/sup> = BC<sup>2<\/sup><br \/>\n\u21d2 \u00a0(x \u2013 1)<sup>2<\/sup> + (y + 1)<sup>2<\/sup> = (3 \u2013 x)<sup>2<\/sup> + (4 \u2013 y)<sup>2<\/sup><br \/>\n\u21d2 \u00a0x<sup>2<\/sup>\u00a0\u2013 2x + 1 + y<sup>2<\/sup> + 2y + 1 = 9 \u2013 6x + x<sup>2<\/sup>\u00a0+ 16 \u2013 8y + y2<br \/>\n\u21d2 \u00a0x<sup>2<\/sup> + y<sup>2<\/sup> \u2013 2x + 2y + 2 = x<sup>2<\/sup> + y<sup>2<\/sup> \u2013 6x \u2013 8y + 25<br \/>\n\u21d2 \u00a04x + 10y = 23<br \/>\n\u21d2 \u00a0x = \u00a0 \\(\\frac { 23-10y }{ 4 }\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(1)<br \/>\nIn right-angled triangle ABC, we have<br \/>\nAB<sup>2<\/sup> + BC<sup>2<\/sup> = AC<sup>2<\/sup><br \/>\n\u21d2 \u00a0(x \u2013 3)<sup>2<\/sup> + (y \u2013 4)<sup>2<\/sup> + (x \u2013 1)<sup>2<\/sup> + (y + 1)<sup>2<\/sup> = (3 \u20131)<sup>2<\/sup> + (4 + 1)<sup>2<\/sup><br \/>\n\u21d2 \u00a0x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup> \u2013 4x \u2013 3y \u2013 1 = 0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;. (2)<br \/>\nSubstituting the value of x from (1) and (2),<br \/>\nwe get<br \/>\n\\({{\\left( \\frac{23-10y}{4} \\right)}^{2}}\\)+ y<sup>2<\/sup> \u2013 (23 \u2013 10y) \u2013 3y \u2013 1 = 0<br \/>\n\u21d2 \u00a04y<sup>2<\/sup> \u2013 12y + 5 = 0 \u00a0\u00a0\u21d2 \u00a0\u00a0(2y \u2013 1) (2y \u2013 5) = 0<br \/>\n\u21d2 \u00a0y = \u00a0\\(\\frac { 1 }{ 2 }\\) \u00a0 or \u00a0\\(\\frac { 5 }{ 2 }\\)<br \/>\nPutting y = \\(\\frac { 1 }{ 2 }\\) \u00a0 and y = \\(\\frac { 5 }{ 2 }\\)respectively in (1) we get<br \/>\nx = \\(\\frac { 9 }{ 2 }\\) \u00a0 and x = \\(\\frac { -1 }{ 2 }\\) \u00a0 \u00a0respectively.<br \/>\nHence, the required vertices of the square are \u00a0\\(\\left( \\frac{9}{2},\\ \\frac{1}{2} \\right)\\text{ \u00a0and \u00a0}\\left( -\\frac{1}{2},\\ \\frac{5}{2} \\right) \\).<\/p>\n<p><strong>Example 11:<\/strong>\u00a0 \u00a0 \u00a0Prove that the points (\u20133, 0), (1, \u20133) and (4, 1) are the vertices of an isosceles right angled triangle. Find the area of this triangle.<br \/>\n<strong>Sol. \u00a0\u00a0<\/strong> Let A (\u20133, 0), B (1, \u20133) and C (4, 1) be the given points. Then,<br \/>\n\\( AB=\\sqrt{{{\\{1-(-3)\\}}^{2}}+{{(-3-0)}^{2}}}=\\sqrt{16+9}=5\\text{ }units. \\)<br \/>\n\\( BC=\\sqrt{{{(4-1)}^{2}}+{{(1+3)}^{2}}}=\\sqrt{9+16}=5\\text{ }units. \\)<br \/>\n\\( CA=\\sqrt{{{(4+3)}^{2}}+{{(1-0)}^{2}}}=\\sqrt{49+1}=5\\sqrt{2}\\text{ }units. \\)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127663\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-11.png\" alt=\"Distance Between Two Points 11\" width=\"253\" height=\"263\" \/><br \/>\nClearly, AB = BC. Therefore, \u2206ABC is isosceles.<br \/>\nAlso, AB<sup>2<\/sup> + BC<sup>2<\/sup> = 25 + 25 = (5)<sup>2<\/sup> = CA<sup>2<\/sup><br \/>\n\u21d2 \u00a0\u2206ABC is right-angled at B.<br \/>\nThus, \u2206ABC is a right-angled isosceles triangle.<br \/>\nNow, Area of \u2206ABC = \\(\\frac { 1 }{ 2 }\\) (Base \u00d7 Height)<br \/>\n= \\(\\frac { 1 }{ 2 }\\) (AB \u00d7 BC)<br \/>\n\u21d2 \u00a0\u00a0Area of \u2206ABC = \\(\\frac { 1 }{ 2 }\\)\u00a0\u00d7 5 \u00d7 5 sq. units<br \/>\n= \\(\\frac { 25 }{ 2 }\\) \u00a0sq. units.<\/p>\n<p><strong>Example 12:<\/strong>\u00a0 \u00a0 \u00a0If P (2, \u2013 1), Q(3, 4), R(\u20132, 3) and S(\u20133, \u20132) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong> The given points are P(2, \u20131), Q(3, 4), R(\u20132, 3) and S(\u20133, \u20132).<br \/>\nWe have,<br \/>\n\\( PQ=\\sqrt{{{(3-2)}^{2}}+{{(4+1)}^{2}}}=\\sqrt{{{1}^{2}}+{{5}^{2}}}=\\sqrt{26}\\text{ }units \\)<br \/>\n\\( QR=\\sqrt{{{(-2-3)}^{2}}+{{(3-4)}^{2}}}=\\sqrt{25+1}=\\sqrt{26}\\text{ }units \\)<br \/>\n\\( RS=\\sqrt{{{(-3+2)}^{2}}+{{(-2-3)}^{2}}}=\\sqrt{1+25}=\\sqrt{26}\\text{ }units \\)<br \/>\n\\( SP=\\sqrt{{{(-3-2)}^{2}}+{{(-2+1)}^{2}}}=\\sqrt{25+1}=\\sqrt{26}\\text{ }units \\)<br \/>\n\\( PR=\\sqrt{{{(-2-2)}^{2}}+{{(3+1)}^{2}}}=\\sqrt{16+16}=4\\sqrt{2}\\text{ }units \\)<br \/>\n\\( QS=\\sqrt{{{(-3-3)}^{2}}+{{(-2-4)}^{2}}}=\\sqrt{36+36}=6\\sqrt{2}\\text{ }units \\)<br \/>\n\u2234 \u00a0PQ = QR = RS = SP = units<br \/>\nand, PR \u2260\u00a0QS<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127664\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-12.png\" alt=\"Distance Between Two Points 12\" width=\"311\" height=\"212\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-12.png 311w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-12-300x205.png 300w\" sizes=\"(max-width: 311px) 100vw, 311px\" \/><br \/>\nThis means that PQRS is a quadrilateral whose sides are equal but diagonals are not equal.<br \/>\nThus, PQRS is a rhombus but not a square.<br \/>\nNow, Area of rhombus PQRS = \\(\\frac { 1 }{ 2 }\\) \u00d7 (Product of lengths of diagonals)<br \/>\n\u21d2 \u00a0Area of rhombus PQRS = \\(\\frac { 1 }{ 2 }\\) \u00d7 (PR \u00d7 QS)<br \/>\n\u21d2 \u00a0Area of rhombus PQRS<br \/>\n\\(=\\left( \\frac{1}{2}\\times 4\\sqrt{2}\\times 6\\sqrt{2} \\right)sq.\\text{ }units\\text{ }=\\text{ }24\\text{ }sq.\\text{ }units\\)<\/p>\n<p><strong>Example 13:<\/strong>\u00a0 \u00a0 \u00a0Find the coordinates of the centre of the circle passing through the points (0, 0), (\u20132, 1) and (\u20133, 2). Also, find its radius.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong> Let P (x, y) be the centre of the circle passing through the points O(0, 0), A(\u20132,1) and B(\u20133,2).<br \/>\nThen, \u00a0OP = AP = BP<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-127665\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2020\/12\/Distance-Between-Two-Points-13.png\" alt=\"Distance Between Two Points 13\" width=\"264\" height=\"227\" \/><br \/>\nNow, OP = AP \u00a0 \u21d2 \u00a0 \u00a0OP<sup>2<\/sup> = AP<sup>2<\/sup><br \/>\n\u21d2 \u00a0x<sup>2<\/sup> + y<sup>2<\/sup> = (x + 2)<sup>2<\/sup> + (y \u2013 1)<sup>2<\/sup><br \/>\n\u21d2 \u00a0x<sup>2<\/sup> + y<sup>2<\/sup> = x<sup>2<\/sup> + y<sup>2<\/sup> + 4x \u2013 2y + 5<br \/>\n\u21d2 \u00a04x \u2013 2y + 5 = 0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(1)<br \/>\nand, OP = BP \u00a0\u21d2 \u00a0 OP<sup>2<\/sup> = BP<sup>2<\/sup><br \/>\n\u21d2 \u00a0x<sup>2<\/sup> + y<sup>2<\/sup> = (x + 3)<sup>2<\/sup> + (y \u2013 2)<sup>2<\/sup><br \/>\n\u21d2 \u00a0x<sup>2<\/sup> + y<sup>2<\/sup> = x<sup>2<\/sup>\u00a0+ y<sup>2<\/sup> + 6x \u2013 4y + 13<br \/>\n\u21d2 \u00a06x \u2013 4y + 13 = 0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(2)<br \/>\nOn solving equations (1) and (2), we get<br \/>\nx = \u00a0\\(\\frac { 3 }{ 2 }\\) \u00a0and \u00a0 y = \u00a0\\(\\frac { 11 }{ 2 }\\)<br \/>\nThus, the coordinates of the centre are \u00a0 \u00a0\\( \\left( \\frac{3}{2},\\ \\frac{11}{2} \\right) \\)<br \/>\n\\(Now,\\text{ }Radius=OP=\\sqrt{{{x}^{2}}+{{y}^{2}}}=\\sqrt{\\frac{9}{4}+\\frac{121}{4}} \\)<br \/>\n\\( =\\frac{1}{2}\\sqrt{130}\\text{ }units. \\)<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Distance Between Two Points Distance Between Two Points Formula (1) Distance between two points on X-axis : The coordinate axes in the coordinate plane can be treated as number lines. If P(x1, 0) and Q(x2, 0) are two points on X-axis, the distance between them is taken as PQ = |x1-x2| &#8230;&#8230;&#8230;.(i) (2) Distance between [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[78,79,80],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Distance Between Two Points - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/distance-between-two-points\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Distance Between Two Points\" \/>\n<meta property=\"og:description\" content=\"Distance Between Two Points Distance Between Two Points Formula (1) Distance between two points on X-axis : The coordinate axes in the coordinate plane can be treated as number lines. 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