Solution: Given, OABC is a square Co-ordinates of A and B are (3, 0) and (p, q) respectively By distance formula, we have

Solution: Given, line equation: x – 4y – 6 = 0 … (i) Co-ordinates of P are (1, 3) Let the co-ordinates of Q be (x , y) Now, the slope of the given line is 4y = x – 6 y = (1/4) x – 6/4 slope (m) = ¼...

(i) Given, co-ordinates of the line joining A ( – 4, 1) and B (17, 10) and point P divides the line segment in the ratio 1 : 2 Let the co- ordinates of P be (x, y) Then,  

Solution: Given, line equations: 4x + 3y = 1 … (1) 5x + 4y = 2 … (2) On solving the above equation to find the point of intersection, we have Multiplying (1) by 4 and (2) by 3 16 x + 12y = 4 15x +...

Solution: Given, ABCD is a rhombus and co-ordinates of A are (3, 6) and of C are (-1, 2) Slope of AC (m1) = (2 – 6)/ (-1 – 3) = -4/-4 = 1 We know that, the diagonals of a rhombus bisect each other...

Solution: Given, points B (1, 3) and D (6, 8) are two opposite vertices of a square ABCD Slope of BD is given by m1 = (8 – 3)/ (6 – 1) = 5/5 = 1 We know that, the diagonal AC is a perpendicular...

(iii) As point (-2, p) lies on the above line The point will satisfy the line equation -2 – 2p + 2 = 0 -2p = 0 p = 0 Thus, the value of p is 0.

Solution: Given, co-ordinates of points A are (7, -3) and of B are (1, 9) (i) The slope of AB (m) = (9 + 3)/ (1 – 7) = 12/ (-6) = -2 (ii) Let PQ be the perpendicular bisector of AB intersecting it...

Solution: The slope of the line joining the points (1, 2) and (5, -6) is m1 = (-6 – 2)/ (5 – 1) = -8/4 = -2 Now, if m2 is the slope of the right bisector of the above line Then, m1 x m2 = -1 -2 x...

  Given, A (2, – 4), B (3, 3) and C (– 1, 5) are the vertices of triangle ABC (i) D is the mid-point of BC So, the co-ordinates of D will be ((3 – 1)/2, (3 + 5)/2) = (2/2, 8/2) = (1, 4) Now,...

Solution: Given, vertices of a triangle are A (10, 4), B (4, – 9) and C (– 2, – 1) Now, Slope of line BC (m1) = (-1 + 9)/ (-2 – 4) = 8/ (-6) = -4/3 Let the slope of the altitude from A (10, 4) to BC...

Solution: The slope of the line joining the points (-3, 2) and (9, 1) is m1 = (1 – 2)/ (9 + 3) = -1/12 Now, let the slope of the line perpendicular to the above line be m2 Then, m1 x m2 = -1 (-1/12)...

From the given figure, its clearly seen that Co-ordinates of A are (2, 3) and of B are (-1, 2) and of C are (3, 0). Now, Slope of BC = (0 – 2)/ (3 – (-1)) = -2/4 = -1/2 So, the slope of the line...

Solution: (i) Here, θ = 45o So, the slope of the line = tan θ = tan 45o = 1 (ii) Equation of the line through P and Q is y – 3 = 1(x – 5) x – y – 2 = 0 (iii) Let the co-ordinates of Q be (0, y)...

Solution: Given, A (– 1, 3), B (4, 2), C (3, – 2) (i) Co-ordinates of centroid G is G (x, y) = ((x1 + x2 + x3)/2, (y1+ y2 + y3)/2) = ((-1 + 4 + 3)/3, (3 + 2 – 2)/3) = (6/3, 3/3) = (2, 1) Hence, the...

Solution: Slope of the line joining the points (0, -2) and (4, 5) is m = (5 + 2)/ (4 – 0) = 7/4 Now, the slope of the line parallel to it and passing through (-1, 3) will be also be 7/4 Hence, the...

Solution: Given, points A (1, 3), B (3, – 1) and C (– 5, – 5) form a triangle Now, Slope of the line AB = m1 = (-1 – 3)/ (3 – 1) = -4/2 = -2 And, Slope of the line BC = m2 = (-5 + 1)/ (-5 – 3) =...

Solution: Let the slope of the line through points (– 2, 6) and (4, 8) be m1 So, m1 = (y2 – y1)/ (x2 – x1) = (8 – 6)/ (4 + 2) = 2/6 = 1/3 And, let the slope of the line through (8, 12) and (4, 24)...

Solution: Let the slope of the line through (0, 0) and (2, 3) be m1 So, m1 = (y2 – y1)/ (x2 – x1) = (3 – 0)/ (2 – 0) = 3/2 And, let the slope of the line through (2, -2) and (6, 4) be m2 So, m2 =...

Solution: Given line equation: 3x + 4y – 7 = 0 (i) Slope of the line is given by, 4y = -3x + 7 y = (-3/4) x + 7 Hence, slope (m1) = -3/4 (ii) Let the slope of the perpendicular to the given line be...

Solution: Given line: 3x + 2y – 8 = 0 2y = -3x + 8 y = (-3/2) x + 4 Here, slope (m1) = -3/2 Now, the co-ordinates of the mid-point of the line segment joining the points (5, -2) and (2, 2) will be...

Solution: Given line: 2x + 5y – 7 = 0 5y = -2x + 7 y = (-2/5) x + 7/5 So, the slope is -2/5 Hence, the slope of the line that is parallel to the given line will be the same, m = -2/5 Now, the...

Solution: Given line: 4x – 3y + 12 = 0 (i) When this line meets the x-axis, its y co-ordinate becomes 0. So, putting y = 0 in the given equation, we get 4x – 3(0) + 12 = 0 4x + 12 = 0 x = -12/4 x =...

Solution: Given line: 3x + 8y = 12 8y = -3x + 12 y = (-3/8) x + 12 So, the slope (m1) = -3/8 Let’s consider the slope of the line perpendicular to the given line as m2 Then, m1 x m2 = -1 -3/8 x m2 =...

Solution: Given line: y = 3x – 5 Here slope (m1) = 3 Substituting x = 0, we get y = – 5 Hence, the y-intercept = – 5 Now, the slope of the line parallel to the given line will be 3 and it passes...

Solution: Given line: 3x + 5y + 15 = 0 5y = -3x – 15 y = (-3/5) x – 3 So, slope (m) = -3/5 The slope of the line parallel to the given line will the same -3/5 And, the line passes through the point...

Solution: Given line: 2x + 5y + 7 = 0 So, its slope is given by 5y = -2x – 7 y = (-2/5) – 7/5 ⇒ m = -2/5 Now, let the slope of the line perpendicular to this line be m’ Then, m x m’ = -1 (-2/5) x m’...

Solution: Given line: 2x – 3y – 7 = 0 Its slope is, 3y = 2x – 7 y = (2/3) x – 7/3 ⇒ m = 2/3 So, the equation of the line parallel to the given line will be 2/3 Also given, the y-intercept is 4 = c...

Solution: Given lines are: 3x + y = 4 … (i) x – ay + 7 = 0 … (ii) bx + 2y + 5 = 0 … (iii) It’s said that these lines form three consecutive sides of a rectangle. So, Lines (i) and (ii) must be...

Solution: Slope of the line passing through the points (-2, 3) and (4, 1) is given by m1 = y2 – y1/ x2 – x1 = (1 – 3)/ (4 + 2) = -2/6 = -1/3 And, the slope of the line: 3x = y + 1 y = 3x -1 Slope...

Solution: Given that the lines 3x + by + 5 = 0 and ax – 5y + 7 = 0 are perpendicular to each other Then the product of their slopes must be -1. Slope of line 3x + by + 5 = 0 is, by = -3x – 5 y =...

Solution: Given, In equation, kx – 5y + 4 = 0 ⇒ 5y = kx + 4 y = (k/5) x + 4/5 So, the slope (m1) = k/5 And, in equation 4x – 2y + 5 = 0 ⇒ 2y = 4x + 5 y = 2x + 5/2 So, the slope (m2) = 2 As the lines...

Solution: Given, Line equation: y/2 = x – p ⇒ y = 2x – 2p Here, the slope of the line is 2. And, another line equation: ax + 5 = 3y ⇒ 3y = ax + 5 y = (a/3) x + 5/3 Hence, the slope of the line is...

Solution: Given lines are: 2x – by + 5 = 0 and ax + 3y = 2 If two lines to be parallel then their slopes must be equal. In equation 2x – by + 5 = 0, by = 2x + 5 y = (2/b) x + 5/b So, the slope of...

Solution: For two lines to be parallel, their slopes must be same. Given line equations, 6x + 5y – 7 = 0 and 2px + 5y + 1 = 0 In equation 6x + 5y – 7 = 0, 5y = -6x + 7 y = (-6/5) x + 7/5 So, the...

Solution: Given straight lines: y = 3x – 5 and 2y = 4x + 7 ⇒ y = 2x + 7/2 And, their slopes are 3 and 2 The product of slopes is 3 x 2 = 6. Hence, as the slopes of both the lines are neither equal...

Solution: Given line equations, 5x + 7y = 3 … (i) 2x – 3 y = 7 … (ii) Now, performing multiplication of (i) by 3 and (ii) by 7, we get 15x + 21y = 9 14x – 21y = 49 On adding we get, 29x = 58 x =...

Solution: Given, The equations of sides of a rectangle are x1 = -1, x2 = 2, y1 = -2, y2 = 6. These lines form a rectangle when they intersect at A, B, C, D respectively Now, The co-ordinates of A,...

(iii) Equation of AB will be y – y1 = m (x – x1) y – (-3) = 3/2 (x – 2) [As P lies on it] y + 3 = 3/2 (x – 2) 2y + 6 = 3x – 6 3x – 2y – 12 = 0

Given, points A and B are on x-axis and y-axis respectively Let co-ordinates of A be (x, 0) and of B be (0, y) And P (2, -3) is the midpoint of AB So, we have 2 = (x + 0)/2 and -3 = (0 + y)/2 x = 4...

(iii) Equation of the side joining A (3, 6) and D (1, -2) is given by 4 (x – 3) = y – 6 4x – 12 = y – 6 4x – y = 6 Thus, the equation of the side joining A (3, 6) and D (1, -2) is 4x – y =...

Solution: Given, the three vertices of a parallelogram ABCD taken in order are A (3, 6), B (5, 10) and C (3, 2) (i) We know that the diagonals of a parallelogram bisect each other. Let (x, y) be the...

Solution: Let the line containing the point P (3, 2) pass through x-axis at A (x, 0) and y-axis at B (0, y) OA = OB Thus, x = y Now, the slope of the line (m) = y2 – y1/ x2 – x1 = 0 – y/ x – 0 =...

Solution: Given, line x – 2y – 11 = 0 passes through y-axis and point (1, 4) So, putting x = 0 in the line equation we get the y-intercept 0 – 2y – 11 = 0 y = -11/2 The co-ordinates are (0, -11/2)...

Solution: Given, P divides the line segment joining the points A (-2, 6) and B (3, -4) in the ratio 2: 3 So, the co-ordinates of P will be x = (m1x2 + m2x1)/(m1 + m2) = (2×3 + 3×(-2))/ (2 + 3) = (6...

Solution: Given, x-intercept of a line is 6 So, The line will pass through the point (6, 0) Also given, the y -intercept of the line is -4 ⇒ c = -4 So, the line will pass through the point (0, -4)...

Given, point (-2, 3) and the x-intercept of the line passing through that point is 4 units. So, the co-ordinates of the point where the line meets the x-axis is (4, 0) Now, slope of the line passing...

Given, ∆ABC and their vertices A (3, 5), B (7, 8) and C (1, – 10). And, AD is median So, D is mid-point of BC Hence, the co-ordinates of D is ([7 + 1]/2, [8 – 10]/2) = (4, -1) Now, Slope of AD, m =...

Solution: (i) Given, ABCD is a parallelogram where A (x, y), B (5, 8), C (4, 7) and D (2, -4) O in the point of intersection of the diagonals of the parallelogram. So, the co-ordinates of O = ([5 +...

Points (h, 4) and (1/2, k) lie on the line passing through A (-1, -1) and B (2, 5) From the graph, its clearly seen that h = 3/2 and k = 2

Solution: Given, Points A (a, 3), B (2,1) and C (5, a) are collinear. So, slope of AB = slope of BC ⇒ -6 = (a – 1) (2 – a) [On cross multiplying] -6 = 2a – 2 – a2 + a -6 = 3a – a2 – 2 a2 – 3a – 4 =...

Solution: Given, two points P (5, 1) and G (1, -1) Slope of the line (m) = y2 – y1/ x2 – x1 = -1 – 1/ 1 – 5 = -2/-4 = ½ So, the equation of the line is y – y1 = m (x – x1) y – 1 = ½ (x – 5) 2y – 2 =...

Solution: Given line equation is 2x – 3y + 12 = 0 On putting y = 0, we will get the intercept made on x-axis 2x – 3y + 12 = 0 2x – 3 × 0 + 12 = 0, 2x – 0 + 2 = 0, 2x = -12 ⇒ x = -6 Now, on putting x...

(iii) It’s seen that the co-ordinates of point of intersection of EF and the x-axis will be y = 0 So, substituting the value y = 0 in the above equation x – y + 4 = 0 x – 0 + 4 = 0 x = -4 Hence, the...

Solution: Gradient of a line (m) = y2 – y1 / x2 – x1 (i) (0, – 2), (3, 4) m = (4 + 2)/(3 – 0) = 6/3 = 2 Hence, gradient = 2 (ii) (3, – 7), (– 1, 8) m = (8 + 7)/(-1 – 3) = 15/-4 Hence, gradient =...

Solution: Given, the equation of straight line passes through (-1, 2) and having slope as 2/5 So, the equation of the line will be y – y1 = m (x – x1) Here, (x1, y1) is (-1, 2) ⇒ y – 2 = (2/5) [x –...

Solution: Given, a line equation passes through point (2, -5) and makes a y-intercept of -3 We know that, The equation of line is y = mx + c, where m is the slope and c is the y-intercept So, we...

Solution: Given, equation of the line is y = mx + c And, it passes through the points (1, 4) So, the point will satisfy the line equation ⇒ 4 = m x 1 + c 4 = m + c m + c = 4 … (i) Also, the line...

Solution: Given, equation of line: y/2 = 3x – 6 And, it passes through the point (a, 2a) Hence, it satisfies the line equation So, 2a/2 = 3(a) – 6 a = 3a – 6 2a = 6 Thus, a =...

Solution: Given, equation of line: y/2 = x – p And, it passes through the point (-4, 4) Hence, it satisfies the line equation So, 4/2 = (-4) – p 2 = -4 – p p = -4 – 2 Thus, p =...

Solution: Given line equations, y = x + 1 and y = √3x – 1 On comparing with y = mx + c, The slope of the line: y = x + 1 is 1 as m = 1 So, tan θ = 1 ⇒ θ = 45o And, The slope of the line: y = √3x – 1...

Solution: Given, equation of line PQ is 3y – 3x + 7 = 0 Re-writing in form of y = mx + c, we have 3y = 3x – 7 ⇒ y = x + (-7/3) Here, (i) Slope = 1 (ii) As tan θ = 1 θ = 45o Hence, the angle which PQ...

(v) y – 3 = 0 y = 3 ⇒ y = (0) x + 3 Hence, slope = 0 and y-intercept = 3 (vi) x – 3 = 0 Here, the slope cannot be defined as the line does not meet y-axis.

(iii) 3x + 5y + 7 = 0 5y = -3x – 7 ⇒ y = (-3/5) x + (-7/5) Hence, slope = -3/5 and y-intercept = -7/5 (iv) x/3 + y/4 = 1 (4x + 3y)/ 12 = 1 4x + 3y = 12 3y = -4x + 12 ⇒ y = (-4/3) x + 4 Hence, slope...

Solution: We know that, equation of line whose slope and y-intercept is given by: y = mx + c, where m is the slope and c is the y-intercept Using the above and converting to this, we find (i) x – 2y...

(iii) Given: gradient = √3, y-intercept = -4/3 ⇒ y = √3x + (-4/3) y = (3√3x – 4)/3 3y = 3√3x – 4 Hence, the equation of line is 3√3x – 3y – 4 = 0. (iv) Given: inclination = 30°, y-intercept = 2...

Solution: Equation of a line whose slope and y-intercept is given by: y = mx + c, where m is the slope and c is the y-intercept (i) Given: slope = 3, y-intercept = – 5 ⇒ y = 3x + (-5) Hence, the...

Solution: The equation of the line parallel to y-axis passing through ( – 3, 5) to x = -3 ⇒ x + 3 = 0

Solution: A line which is parallel to y-axis is x = a (i) Here, x = 3 Hence, the equation of line parallel to y-axis is at a distance of 3 units to the right is x – 3 = 0. (ii) Here, x = -2 Hence,...

Solution: (i) A line which is parallel to x-axis is y = a ⇒ y = 2 Hence, the equation of line parallel to x-axis which is at a distance of 2 units above it is y – 2 = 0. (ii) A line which is...

(iii) tan θ = 1/√3 ⇒ θ = 30o