CBSE Class 10 Coordinate geometry Sure Shot Questions Set 05

Question. Find the ratio in which the line \( 3x + 4y - 9 = 0 \) divides the line segment joining the points \( (1, 3) \) and \( (2, 7) \). 
Answer: Let the ratio be \( k : 1 \).
\(\therefore\) Coordinates of R are:
\( \left( \frac{2k + 1}{k + 1}, \frac{7k + 3}{k + 1} \right) \),
Since, R lies on the line \( 3x + 4y - 9 = 0 \)
\(\therefore 3 \left( \frac{2k + 1}{k + 1} \right) + 4 \left( \frac{7k + 3}{k + 1} \right) - 9 = 0 \)
\(\Rightarrow 6k + 3 + 28k + 12 - 9k - 9 = 0 \)
\(\Rightarrow (6k + 28k - 9k) + (3 + 12 - 9) = 0 \)
\(\Rightarrow 25k + 6 = 0 \)
\(\Rightarrow k = -\frac{6}{25} \)
\(\therefore\) The required ratio is \( -6 : 25 \) or \( 6 : 25 \) (externally).

Question. If the point \( P(x, y) \) is equidistant from the points \( A(3, 6) \) and \( B(-3, 4) \), prove that \( 3x + y - 5 = 0 \). 
Answer: \(\because P \) is equidistant from A and B.
\(\therefore AP = BP \)
\(\Rightarrow AP^2 = BP^2 \)
\(\Rightarrow (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 \)
\(\Rightarrow x^2 - 6x + 9 + y^2 - 12y + 36 = x^2 + 6x + 9 + y^2 - 8y + 16 \)
\(\Rightarrow x^2 + y^2 - 6x - 12y + 45 = x^2 + y^2 + 6x - 8y + 25 \)
\(\Rightarrow (-6x - 6x) + (-12y + 8y) + 45 - 25 = 0 \)
\(\Rightarrow -12x + 20 - 4y = 0 \)
\(\Rightarrow -3x - y + 5 = 0 \)
or \( 3x + y - 5 = 0 \)

Question. The coordinates of A and B are \( (1, 2) \) and \( (2, 3) \). If P lies on AB, then find the coordinates of P such that: \( \frac{AP}{PB} = \frac{4}{3} \). 
Answer: \(\because \frac{AP}{PB} = \frac{4}{3} \therefore AP : PB = 4 : 3 \)
Here, P divides AB internally in the ratio \( 4 : 3 \).
\(\therefore\) P has coordinates as:
\( \left[ \frac{4 \times 2 + 3 \times 1}{4 + 3}, \frac{4 \times 3 + 3 \times 2}{4 + 3} \right] \),
or \( \left[ \frac{8 + 3}{7}, \frac{12 + 6}{7} \right] \),
or \( \left[ \frac{11}{7}, \frac{18}{7} \right] \)

Question. Show that the triangle PQR formed by the points \( P(\sqrt{2}, \sqrt{2}), Q(-\sqrt{2}, -\sqrt{2}) \) and \( R(-\sqrt{6}, \sqrt{6}) \) is an equilateral triangle. 
Answer: We have, \( P(\sqrt{2}, \sqrt{2}), Q(-\sqrt{2}, -\sqrt{2}) \) and \( R(-\sqrt{6}, \sqrt{6}) \)
\(\therefore PQ = \sqrt{(-\sqrt{2} - \sqrt{2})^2 + (-\sqrt{2} - \sqrt{2})^2} = \sqrt{(-2\sqrt{2})^2 + (-2\sqrt{2})^2} \)
\( = \sqrt{4 \times 2 + 4 \times 2} = \sqrt{8 + 8} = \sqrt{16} = 4 \)
\( PR = \sqrt{(-\sqrt{6} - \sqrt{2})^2 + (\sqrt{6} - \sqrt{2})^2} \)
\( = \sqrt{6 + 2 + 2\sqrt{12} + 6 + 2 - 2\sqrt{12}} = \sqrt{8 + 8} = \sqrt{16} = 4 \)
\( RQ = \sqrt{(-\sqrt{2} - (-\sqrt{6}))^2 + (-\sqrt{2} - \sqrt{6})^2} \)
\( = \sqrt{2 + 6 - 2\sqrt{12} + 2 + 6 + 2\sqrt{12}} = \sqrt{8 + 8} = \sqrt{16} = 4 \)
Since, \( PQ = PR = RQ = 4 \)
\(\therefore PQR \) is an equilateral triangle.

Question. The line joining the points \( (2, -1) \) and \( (5, -6) \) is bisected at P. If P lies on the line \( 2x + 4y + k = 0 \), find the value of k.
Answer: We have \( A(2, -1) \) and \( B(5, -6) \).
\(\because P \) is the mid point of AB,
\(\therefore\) Coordinates of P are: \( \left[ \frac{2 + 5}{2}, \frac{-1 - 6}{2} \right] \) or \( \left( \frac{7}{2}, \frac{-7}{2} \right) \)
Since P lies on the line \( 2x + 4y + k = 0 \)
\(\therefore\) We have:
\( 2 \left( \frac{7}{2} \right) + 4 \left( \frac{-7}{2} \right) + k = 0 \)
\(\Rightarrow 7 - 14 + k = 0 \)
\(\Rightarrow -7 + k = 0 \Rightarrow k = 7 \)

Question. Find the point on y-axis which is equidistant from the points \( (5, -2) \) and \( (-3, 2) \). 
Answer: \(\because\) Let P is on the y-axis \(\therefore\) Coordinates of P are: \( (0, y) \)
Since, \( PA = PB \therefore PA^2 = PB^2 \)
\(\Rightarrow (5 - 0)^2 + (-2 - y)^2 = (-3 - 0)^2 + (2 - y)^2 \)
\(\Rightarrow 25 + 4 + 4y + y^2 = 9 + 4 - 4y + y^2 \)
\(\Rightarrow 25 + 4y = 9 - 4y \)
\(\Rightarrow 8y = -16 \)
\(\Rightarrow y = -2 \)
\(\therefore\) The required point is \( (0, -2) \).

Question. The line joining the points \( (2, 1) \) and \( (5, -8) \) is trisected at the points P and Q. If point P lies on the line \( 2x - y + k = 0 \), find the value of k. 
Answer: \(\because AB \) is trisected at P and Q \(\therefore\) Coordinates of P are:
\( \left[ \frac{1 \times 5 + 2 \times 2}{1 + 2}, \frac{1 \times (-8) + 2 \times 1}{1 + 2} \right] \)
or \( \left[ \frac{9}{3}, \frac{-6}{3} \right] \) or \( (3, -2) \)
Since, \( P(3, -2) \) lies on \( 2x - y + k = 0 \)
\(\therefore\) We have:
\( 2(3) - (-2) + k = 0 \)
\(\Rightarrow 6 + 2 + k = 0 \)
\(\Rightarrow 8 + k = 0 \Rightarrow k = -8 \)

Question. If \( P(x, y) \) is any point on the line joining the points \( A(a, 0) \) and \( B(0, b) \), then show that: \( \frac{x}{a} + \frac{y}{b} = 1 \). 
Answer: \(\because P \) lies on the line joining A and B.
\(\therefore A, B \) and P are collinear.
\(\Rightarrow\) The area of a \( \Delta \) formed by \( A(a, 0), B(0, b) \) and \( P(x, y) \) is zero.
\(\therefore\) We have:
\( x_1[y_2 - y_3] + x_2[y_3 - y_1] + x_3[y_1 - y_2] = 0 \)
\(\Rightarrow x[0 - b] + a[b - y] + 0[y - 0] = 0 \)
\(\Rightarrow -bx + ab - ay = 0 \)
\(\Rightarrow -(bx + ay) = -ab \)
\(\Rightarrow bx + ay = ab \)
\(\Rightarrow \frac{bx}{ab} + \frac{ay}{ab} = \frac{ab}{ab} \) [Dividing by ab]
\(\Rightarrow \frac{x}{a} + \frac{y}{b} = 1 \)

Question. Find the point on x-axis which is equidistant from the points \( (2, -5) \) and \( (-2, 9) \). 
Answer: \(\because\) The required point ‘P’ is on x-axis. \(\therefore\) Coordinates of P are \( (x, 0) \).
\(\therefore\) We have \( AP = PB \Rightarrow AP^2 = PB^2 \)
\(\Rightarrow (2 - x)^2 + (-5 + 0)^2 = (-2 - x)^2 + (9 - 0)^2 \)
\(\Rightarrow 4 - 4x + x^2 + 25 = 4 + 4x + x^2 + 81 \)
\(\Rightarrow -4x + 25 = 4x + 81 \)
\(\Rightarrow -8x = 56 \)
\(\Rightarrow x = \frac{56}{-8} = -7 \)
\(\therefore\) The required point is \( (-7, 0) \).

Question. The line segment joining the points \( P(3, 3) \) and \( Q(6, -6) \) is trisected at the points A and B such that A is nearer to P. It also lies on the line given by \( 2x + y + k = 0 \). Find the value of k. 
Answer: \(\because PQ \) is trisected by A such that \( PA : AQ = 1 : 2 \)
\(\therefore\) The coordinates of A are:
\( \left[ \frac{1 \times 6 + 2 \times 3}{1 + 2}, \frac{1 \times (-6) + 2 \times 3}{1 + 2} \right] \),
or \( \left[ \frac{6 + 6}{3}, \frac{-6 + 6}{3} \right] \),
or \( \left[ \frac{12}{3}, \frac{0}{3} \right] \) or \( (4, 0) \).
Since, \( A(4, 0) \) lies on the line \( 2x + y + k = 0 \)
\(\therefore 2(4) + (0) + k = 0 \)
\(\Rightarrow 8 + k = 0 \Rightarrow k = -8 \)

Question. Find the ratio in which the points \( (2, 4) \) divides the line segment joining the points \( A(-2, 2) \) and \( B(3, 7) \). Also find the value of y. 
Answer: Let \( P(2, y) \) divides the join of \( A(-2, 2) \) and \( B(3, 7) \) in the ratio \( k:1 \)
\(\therefore\) Coordinates of P are: \( \left[ \frac{3k - 2}{k + 1}, \frac{7k + 2}{k + 1} \right] \),
\(\Rightarrow \frac{3k - 2}{k + 1} = 2 \) and \( \frac{7k + 2}{k + 1} = y \)
Now, \( \frac{3k - 2}{k + 1} = 2 \Rightarrow 3k - 2 = 2k + 2 \Rightarrow k = 4 \)
And \( y = \frac{7k + 2}{k + 1} \Rightarrow y = \frac{7(4) + 2}{4 + 1} \)
\(\Rightarrow y = \frac{30}{5} = 6 \)
Thus, \( y = 6 \) and \( k = 4 \)

Question. Find the area of the triangle formed by joining the mid-points of the sides of triangle whose vertices are \( (0, - 1), (2, 1) \), and \( (0, 3) \). 
Answer: We have the vertices of the given triangle as \( A (0, - 1), B (2, 1) \) and \( C (0, 3) \). Let D, E and F be the mid-points of \( AB, BC \) and \( AC \).
\(\therefore\) Coordinates of D are \( \left[ \frac{0 + 2}{2}, \frac{- 1 + 1}{2} \right] \) or \( (1, 0) \)
E are \( \left[ \frac{2 + 0}{2}, \frac{1 + 3}{2} \right] \) or \( (1, 2) \)
F are \( \left[ \frac{0 + 0}{2}, \frac{3 + (- 1)}{2} \right] \) or \( (0, 1) \)
\(\therefore\) Coordinates of the vertices of \( \Delta DEF \) are \( (1, 0), (1, 2) \) and \( (0, 1) \).
Now, area of \( \Delta DEF = \frac{1}{2} [1 (2 - 1) + 1 (1 - 0) + 0 (0 - 2)] = \frac{1}{2} \times 2 = 1 \) sq. units.

Question. Find the ratio in which the point \( (x, - 1) \) divides the line segment joining the points \( (- 3, 5) \) and \( (2, - 5) \). Also find the value of \( x \). 
Answer: Let the required ratio is \( k : 1 \)
\(\therefore\) The coordinates of \( P \) are: \( \left[ \frac{2k - 3}{k + 1}, \frac{- 5k + 5}{k + 1} \right] \),
But the coordinates of \( P \) are \( (x, - 1) \)
\(\therefore \frac{- 5k + 5}{k + 1} = - 1 \Rightarrow - 5k + 5 = - k - 1 \Rightarrow 2k = 3 \) or \( k = \frac{3}{2} \)
Also, \( x = \frac{2k - 3}{k + 1} = \frac{2(\frac{3}{2}) - 3}{\frac{3}{2} + 1} = \frac{3 - 3}{\frac{5}{2}} = 0 \)
\(\therefore x = 0 \) and \( k = \frac{3}{2} \)

Question. If the mid-point of the line segment joining the point \( A(3, 4) \) and \( B(k, 6) \) is \( P(x, y) \) and \( x + y - 10 = 0 \), then find the value of \( k \). 
Answer: \( \because \) Mid point of the line segment joining \( A(3, 4) \) and \( B(k, 6) = \left( \frac{3 + k}{2}, \frac{4 + 6}{2} \right) = \left( \frac{3 + k}{2}, 5 \right) \)
\(\therefore \frac{3 + k}{2} = x \) and \( 5 = y \)
Since, \( x + y - 10 = 0 \)
\(\Rightarrow \frac{3 + k}{2} + 5 - 10 = 0 \)
\(\Rightarrow 3 + k + 10 - 20 = 0 \)
\(\Rightarrow 3 + k = 10 \)
\(\Rightarrow k = 10 - 3 = 7 \)
Thus, the required value of \( k = 7 \)

Question. Point \( P, Q, R \) and \( S \) divide the line segment joining the points \( A (1, 2) \) and \( B (6, 7) \) in \( 5 \) equal parts. Find the co-ordinates of the points \( P, Q \) and \( R \). 
Answer: \( \therefore P, Q, R \) and \( S \), divide \( AB \) into five equal parts.
\( \therefore AP = PQ = QR = RS = SB \)
Now, \( P \) divides \( AB \) in the ratio \( 1 : 4 \)
Let, the co-ordinates of \( P \) be \( x \) and \( y \).
\(\therefore\) Using the section formula i.e., \( x = \frac{mx_2 + nx_1}{m + n} \), \( y = \frac{my_2 + ny_1}{m + n} \), we have
\(\therefore x = \frac{1(6) + 4(1)}{1 + 4} = \frac{6 + 4}{5} = 2 \)
\( y = \frac{1(7) + 4(2)}{1 + 4} = \frac{7 + 8}{5} = 3 \)
\( (x, y) = (2, 3) \)
Next, \( Q \) divides \( AB \) in the ratio \( 2 : 3 \)
\(\therefore\) Co-ordinates of \( Q \) are : \( \left[ \frac{2(6) + 3(1)}{2 + 3}, \frac{2(7) + 3(2)}{2 + 3} \right] \) or \( \left[ \frac{15}{5}, \frac{20}{5} \right] \) or \( (3, 4) \)
Now, \( R \) divides \( AB \) in the ratio \( 3 : 2 \)
\(\Rightarrow\) Co-ordinates of \( R \) are : \( \left[ \frac{3(6) + 2(1)}{3 + 2}, \frac{3(7) + 2(2)}{3 + 2} \right] \) or \( \left[ \frac{20}{5}, \frac{25}{5} \right] \) or \( (4, 5) \)
The co-ordinates of \( P, Q \) and \( R \) are respectively : \( (2, 3), (3, 4) \) and \( (4, 5) \).

Question. The line-segment joining the points \( (3, - 4) \) and \( (1, 2) \) is trisected at the points \( P \) and \( Q \). If the coordinates of \( P \) and \( Q \) are \( (p, - 2) \) and \( \left( \frac{5}{3}, q \right) \) respectively, find the values of \( p \) and \( q \). 
Answer: \( p = \frac{7}{3}; q = 0 \)

Question. The line joining the points \( (2, 1) \) and \( (5, - 8) \) is trisected at the points \( P \) and \( Q \). If \( P \) lies on the line \( 2x - y + k = 0 \), find the value of \( k \). 
Answer: \( k = -8 \) or \( k = -13 \)

Question. If the coordinates of the mid-points of the sides of a \( \Delta \) are \( (10, 5), (8, 4) \) and \( (6, 6) \), then find the coordinates of its vertices. 
Answer: \( (4, 5), (8, 3) \) and \( (12, 3) \)

Question. Find the coordinates of the points which divide the line segment joining the points \( (- 4, 0) \), and \( (0, 6) \) in three equal pasts.
Answer: \( \left( -\frac{8}{3}, 2 \right), \left( \frac{-4}{3}, 3 \right) \)

Question. Find the coordinates of the point equidistant from the points \( A (1, 2), B (3, - 4) \) and \( C (5, - 6) \). 
Answer: \( (-1, -2) \)

Question. Prove that the points \( A (- 4, - 1), B (- 2, - 4), C (4, 0) \) and \( D (2, 3) \) are the vertices of a rectangle. 
Answer: [Proof required]

Question. Find the coordinates of the points which divide the line-segment joining the points \( (- 4, 0) \) and \( (0, 6) \) in four equal parts.
Answer: \( \left( -3, \frac{3}{2} \right), (2, 3) \) and \( \left( -1, \frac{9}{2} \right) \)

Question. Find the coordinates of the points which divide the line-segment joining the points \( (2, -2) \) and \( (-7, 4) \) in three equal parts. 
Answer: \( (-1, 0) \) and \( (-4, 2) \)

Question. Find the coordinates of the point equidistant from the points \( A (5, 1), B (- 3, -7) \) and \( C (7, - 1) \). 
Answer: \( (2, -4) \)

Question. The vertices of a \( \Delta ABC \) and given by \( A (2, 3) \) and \( B (- 2, 1) \) and its centroid is \( G \left( 1, \frac{2}{3} \right) \). Find the coordinates of the third vertex \( C \) of the \( \Delta ABC \). 
Answer: \( (3, -2) \)

Question. If the points \( (x, y) \) is equidistant from the points \( (a + b, b - a) \) and \( (a - b, a + b) \). Prove that \( bx = ay \). 
Answer: [Proof required]

Question. Two vertices of a \( \Delta ABC \) are given by \( A (6, 3) \) and \( B (–1, 7) \) and its centroid is \( G (1, 5) \). Find the coordinates of the third vertex of the \( \Delta ABC \). 
Answer: \( (–2, 5) \)

Question. Two of the vertices of a \( \Delta ABC \) are given by \( A (6, 4) \) and \( B (- 2, 2) \) and its centroid is \( G (3, 4) \). Find the coordinates of the third vertex \( C \) of the \( \Delta ABC \). 
Answer: \( (5, 6) \)

Question. The coordinates one end point of a diameter of a circle are \( (4, -1) \). If the coordinates of the centre be \( (1, -3) \) find the coordinates of the other end of the diameter. 
Answer: \( (–2, –5) \)

Question. Show that the points \( A (1, 2), B (5, 4) C (3, 8) \) and \( D (- 1, 6) \) are the vertices of a square. 
Answer: [Proof required]

Question. Find the value of \( P \) for which the points \( (- 1, 3), (2, p) \) and \( (5, - 1) \) are collinear. 
Answer: \( p = 1 \)

Question. Find the distance of the point \( (– 6, 8) \) from the origin. 
Answer: \( 10 \)

Question. Find the coordinates of the point equidistant from three given points \( A (5, 3), B (5, - 5) \) and \( C (1, - 5) \).
Answer: \( (3, –1) \)

Question. Find the value of \( p \) for which the points \( (- 5, 1), (1, p) \) and \( (4, - 2) \) are collinear. 
Answer: \( p = –1 \)

Question. Find the coordinates of the point on the line joining \( P (1, - 2) \) and \( Q (4, 7) \) that is twice as far from \( P \) as from \( Q \). 
Answer: \( (2, 1) \)

Question. Find the perimeter of a triangle with vertices \( (0, 4), (0, 0) \) and \( (3, 0) \).
Answer: \( 12 \) units

Question. In what ratio is the line segment joining the points \( (- 2, - 3) \) and \( (3, 7) \) divided by the \( y \)–axis? Also, find the coordinates of the point of division. 
Answer: \( 2 : 3, (0, 1) \)

Question. If \( A (5, - 1), B (- 3, - 2) \) and \( C (- 1, 8) \) are the vertices of \( \Delta ABC \), find the length of median through \( A \) and the coordinates of the centroid. 
Answer: \( \sqrt{65}; \left( \frac{1}{3}, \frac{5}{3} \right) \)

Question. Find the value of \( k \) if the points \( A (2, 3), B (4, k) \) and \( C (6, - 3) \) are collinear. 
Answer: \( k = 0 \)

Question. If \( (- 2, - 1), (a, 0), (4, b) \) and \( (1, 2) \) are the vertices of a parallelogram, find the value of \( ‘a’ \) and \( ‘b’ \).
Answer: \( a = 1, b = 3 \)

Question. The vertices of a triangle are \( (- 1, 3), (1, - 1) \) and \( (5, 1) \). Find the lengths of medians through vertices \( (- 1, 3) \) and \( (5, 1) \). 
Answer: \( \left( \frac{5}{3}, 1 \right) \)

Question. In what ratio the line \( x - y - 2 = 0 \) divides the line segment joining \( (3, - 1) \) and \( (8, 9) \)? 
Answer: \( 2 : 3 \)

Question. Find the ratio in which the line joining the points \( (6, 4) \) and \( (1, - 7) \) is divided by \( x \)–axis. 
Answer: \( 4 : 7 \)

Question. Find the ratio in which the point \( (- 3, k) \) divides the line segment joining the points \( (- 5, - 4) \) and \( (- 2, 3) \). Hence find the value of \( k \). 
Answer: \( 2 : 1, k = \frac{2}{3} \)

Question. Three consecutive vertices of a parallelogram are \( (- 2, - 1), (1, 0) \) and \( (4, 3) \). Find the coordinates of the fourth vertex. 
Answer: \( (1, 2), (1, 4) \)

Question. For what value \( P \), are the points \( (2, 1), (p, - 1) \) and \( (- 1, 3) \) collinear?
Answer: \( p = 5 \)

Question. Find the value (s) of \( k \) for which the points \( [(3k – 1), (k – 2)], [k, (k – 7)] \) and \( [(k – 1), (– k – 2)] \) are collinear.
Answer: \( k = 0, 3 \)

Question. If the point \( A (0, 2) \) is equidistant from the point \( B(3, p) \) and \( C (p, 5) \), find \( p \).
Answer: \( p = 1 \)

Question. Find the ratio in which the point \( P(x, 2) \) divides the line segment joining the points \( A(12, 5) \) and \( B(4, – 3) \). Also find the value of \( x \). 
Answer: \( x = 9 \) and ratio \( = 3 : 5 \)

Question. If the point \( P(k – 1, 2) \) is equidistant from the point \( A (3, k) \) and \( B(k, 5) \) then find the values of \( k \). 
Answer: \( k = 5 \)

Question. Find the ratio in which the line segment joining the points \( A(3, –3) \) and \( B(–2, 7) \) is divided by \( x \)– axis. Also find the co-ordinates of the point of division. 
Answer: \( x = \frac{3}{2} \) and Required ratio as \( 7 : 3 \); co-ordinates of the point of division \( \left( \frac{3}{2}, 0 \right) \)

Question. The mid-point \( P \) of the line segment joining the points \( A(–10, 4) \) and \( B(–2, 0) \) lies on the line segment joining the points \( C(–9, –4) \) and \( D(–4, y) \). Find the ratio in which \( P \) divides \( CD \). Also find the value of \( y \).
Answer: Required ratio \( = 3 : 2 \) and \( y = 6 \).