{"id":31851,"date":"2018-10-08T11:51:19","date_gmt":"2018-10-08T11:51:19","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=31851"},"modified":"2020-11-25T14:51:41","modified_gmt":"2020-11-25T09:21:41","slug":"ml-aggarwal-class-10-solutions-for-icse-maths-chapter-12-chapter-test","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/ml-aggarwal-class-10-solutions-for-icse-maths-chapter-12-chapter-test\/","title":{"rendered":"ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test"},"content":{"rendered":"
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths<\/a>. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 12 Equation of a Straight Line Chapter Test.<\/p>\n ML Aggarwal Solutions<\/a>ICSE Solutions<\/a>Selina ICSE Solutions<\/a><\/p>\n Question 1.<\/strong><\/span> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Question 14.<\/strong><\/span> Question 15.<\/strong><\/span> Question 16.<\/strong><\/span>
\nFind the equation of a line whose inclination is 60\u00b0 and y-intercept is – 4.<\/strong>
\nSolution:<\/strong><\/span>
\nAngle of inclination = 60\u00b0
\nSlope = tan \u03b8 = tan 60\u00b0 = \u221a3
\nEquation of the line will be,
\ny = mx + c = \u221a3x + ( – 4)
\n=> y – \u221a3x – 4 Ans.<\/p>\n
\nWrite down the gradient and the intercept on the y-axis of the line 3y + 2x = 12.<\/strong>
\nSolution:<\/strong><\/span>
\nSlope of the line 3y + 2x = 12
\n=> 3y = 12 – 2x
\n=> 3y = – 2x + 12
\n<\/p>\n
\nIf the equation of a line is y – \u221a3x + 1, find its inclination.<\/strong>
\nSolution:<\/strong><\/span>
\nIn the line
\ny = \u221a3 x + 1
\nSlope = \u221a3 => tan \u03b8 = \u221a3
\n\u03b8 = 60\u00b0 (\u2235 tan 60\u00b0 = \u221a3)<\/p>\n
\nIf the line y = mx + c passes through the points (2, – 4) and ( – 3, 1), determine the values of m and c.<\/strong>
\nSolution:<\/strong><\/span>
\nThe equation of line y = mx + c
\n\u2235 it passes through (2, – 4) and ( – 3, 1)
\nNow substituting the value of these points – 4 = 2 m + c …(i)
\nand 1 = – 3 m + c …(ii)
\nSubtracting we get,
\n<\/p>\n
\nIf the point (1, 4), (3, – 2) and (p, – 5) lie on a st. line, find the value of p.<\/strong>
\nSolution:<\/strong><\/span>
\nLet the points to be A (1, 4), B (3, – 2) and C (p, – 5) are collinear and let B (3, – 2)
\ndivides AC in the ratio of m1 : m2
\n<\/p>\n
\nFind the inclination of the line joining the points P (4, 0) and Q (7, 3).<\/strong>
\nSolution:<\/strong><\/span>
\nSlope of the line joining the points P (4, 0) and Q (7, 3)
\n<\/p>\n
\nFind the equation of the line passing through the point of intersection of the lines 2x + y = 5 and x – 2y = 5 and having y-intercept equal to \\(– \\frac { 3 }{ 7 } \\)<\/strong>
\nSolution:<\/strong><\/span>
\nEquation of lines are
\n2x + y = 5 …(i)
\nx – 2y = 5 …(ii)
\nMultiply (i) by 2 and (ii) by 1, we get
\n4x + 2y = 10
\nx – 2y = 5
\nAdding we get,
\n<\/p>\n
\nIf the point A is reflected in the y-axis, the co-ordinates of its image A1<\/sub>, are (4, – 3),<\/strong>
\n(i) Find the co-ordinates of A<\/strong>
\n(ii) Find the co-ordinates of A2<\/sub>, A3<\/sub> the images of the points A, A1<\/sub>, Respectively under reflection in the line x = – 2<\/strong>
\nSolution:<\/strong><\/span>
\n(i) \u2235 A is reflected in the y-axis and its image is A1<\/sub> (4, – 3)
\nCo-ordinates of A will be ( – 4, – 3)
\n
\n<\/p>\n
\nIf the lines \\(\\frac { x }{ 3 } +\\frac { y }{ 4 } =7 \\) and 3x + ky = 11 are perpendicular to each other, find the value of k.<\/strong>
\nSolution:<\/strong><\/span>
\nGiven
\nEquation of lines are
\n<\/p>\n
\nWrite down the equation of a line parallel to x – 2y + 8 = 0 and passing through the point (1, 2).<\/strong>
\nSolution:<\/strong><\/span>
\nThe equation of the line is x – 2y + 8 = 0
\n=> 2y = x + 8
\n<\/p>\n
\nWrite down the equation of the line passing through ( – 3, 2) and perpendicular to the line 3y = 5 – x.<\/strong>
\nSolution:<\/strong><\/span>
\nEquations of the line is
\n3y = 5 – x
\n=> 3y = – x + 5
\n<\/p>\n
\nFind the equation of the line perpendicular to the line joining the points A (1, 2) and B (6, 7) and passing through the point which divides the line segment AB in the ratio 3 : 2.<\/strong>
\nSolution:<\/strong><\/span>
\nLet slope of the line joining the points A (1, 2) and B (6, 7) be m1<\/sub>
\n
\n<\/p>\n
\nThe points A (7, 3) and C (0, – 4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.<\/strong>
\nSolution:<\/strong><\/span>
\nSlope of line AC (m1<\/sub>)
\n
\n<\/p>\n
\nA straight line passes through P (2, 1) and cuts the axes in points A, B. If BP : PA = 3 : 1, find:<\/strong>
\n
\n(i) the co-ordinates of A and B<\/strong>
\n(ii) the equation of the line AB<\/strong>
\nSolution:<\/strong><\/span>
\nA lies on x-axis and B lies on y-axis
\nLet co-ordinates of A be (x, 0) and B be (0, y) , and P (2, 1) divides BA in the ratio 3 : 1.
\n
\n<\/p>\n
\nA straight line makes on the co-ordinates axes positive intercepts whose sum is 7. If the line passes through the point ( – 3, 8), find its equation.<\/strong>
\nSolution:<\/strong><\/span>
\nLet the line make intercept a and b with the x-axis and y-axis respectively then the line passes through
\n
\n<\/p>\n
\nIf the coordinates of the vertex A of a square ABCD are (3, – 2) and the quation of diagonal BD is 3 x – 7 y + 6 = 0, find the equation of the diagonal AC. Also find the co-ordinates of the centre of the square.<\/strong>
\nSolution:<\/strong><\/span>
\nCo-ordinates of A are (3, – 2).
\n
\n
\n
\n<\/p>\n