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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Additional Questions

I. Multiple Choice Questions

Question 1.

If the three points (-3, 7), (a, 1), (-3, 2) are collinear then the value of “a” is

(1) 0

(2) -1

(3) -3

(4) 1

Answer:

(3) -3

Hint:

Since the three points are collinear

Area of a ∆ = 0

-3 + 2a – 21 – (7a – 3 – 6) = 0 ⇒ 2a – 24 – 7a + 9 = 0

– 5a – 15 = 0 ⇒ – 5(a + 3) = 0

a + 3 = 0 ⇒ a = -3

Question 2.

If A (5, 5), B (-5, 1), C (10, 7) lie in a straight line, then the area of ∆ ABC is …………….

(1) \(\frac { 13 }{ 2 } \) sq.units

(2) 9 sq.units

(3) 25 sq.units

(4) 0

Answer:

(4) 0

Hint:

Area of the ∆^{le
}

Question 3.

In a rectangle ABCD, area of ∆ ABC is \(\frac { 31 }{ 2 } \) sq. units. Then the area of rectangle is ……………

(1) 62 sq. units

(2) 31 sq. units

(3) 60 sq. units

(4) 30 sq. units

Answer:

(2) 31 sq. units

Hint:

In a rectangle area of ∆ ABC and area of ∆ ACD are equal.

Area of rectangle ABCD = 2 × \(\frac { 31 }{ 2 } \) = 31 sq.units

Question 4.

If the points (k, 2k), (3k, 3k) and (3,1) are collinear, then k is ……………..

(1) \(\frac { 1 }{ 3 } \)

(2) – \(\frac { 1 }{ 3 } \)

(3) \(\frac { 2 }{ 3 } \)

(4) – \(\frac { 2 }{ 3 } \)

Answer:

(2) – \(\frac { 1 }{ 3 } \)

Hint:

Since the three points are collinear. Area of a ∆ = 0

3k^{2} + 3k + 6k – (6k^{2} + 9k + k) = 0 ⇒ 3k^{2} + 9k – 6k^{2} – 10k = 0

-3 k^{2} – k = 0 ⇒ -k(3k + 1) = 0

3k + 1 = 0 ⇒ 3 k = -1 ⇒ k = – \(\frac { 1 }{ 3 } \)

Question 5.

If the area of the triangle formed by the points (x, 2x), (-2, 6) and (3, 1) is 5 square units then x = ………….

(1) 2

(2) \(\frac { 3 }{ 5 } \)

(3) 3

(4) 5

Answer:

(1) 2

Hint:

Area of the triangle = 5 sq. units

6x – 2 + 6x – (-4x + 18 + x) = 10 ⇒ 12x – 2 – (-3x + 18) = 10

12x – 2 + 3x – 18 = 10

15x – 20 = 10 ⇒ 15x = 10 + 20 = 30

x = \(\frac { 30 }{ 15 } \) = 2

Question 6.

The slope of a line parallel to y-axis is equal to …………..

(1) 0

(2) -1

(3) 1

(4) not defined

Answer:

(4) not defined

Question 7.

In a rectangle PQRS, the slope of PQ = \(\frac { 5 }{ 6 } \) then the slope of RS is ………..

(1) \(\frac { -5 }{ 6 } \)

(2) \(\frac { 6 }{ 5 } \)

(3) \(\frac { -6 }{ 5 } \)

(4) \(\frac { 5 }{ 6 } \)

Answer:

\(\frac { 5 }{ 6 } \)

Hint:

In a rectangle opposite sides are parallel.

∴ Slope of the line RS is \(\frac { 5 }{ 6 } \).

Question 8.

The y – intercept of the line y = 2x is ………

(1) 1

(2) 2

(3) \(\frac { 1 }{ 2 } \)

(4) 0

Answer:

(4) 0

Question 9.

The straight line given by the equation y = 5 is …………..

(1) Parallel to x – axis

(2) Parallel to y – axis

(3) Passes through the origin

(4) None of these

Answer:

(1) Parallel to x – axis

Question 10.

The x – intercept of the line 2x – 3y + 5 = 0 is ………….

(1) \(\frac { 5 }{ 2 } \)

(2) \(\frac { -5 }{ 2 } \)

(3) \(\frac { 2 }{ 5 } \)

(4) \(\frac { -2 }{ 5 } \)

Answer:

(2) \(\frac { -5 }{ 2 } \)

Hint:

2x – 3y + 5 = 0 ⇒ 2x – 3y = – 5

Question 11.

The lines 3x – 5y + 1 = 0 and 5x + ky + 2 = 0 are perpendicular if the value of k is ………..

(1) -5

(2) 3

(3) -3

(4) 5

Answer:

(2) 3

Hint:

Slope of the first line (m_{1}) = \(\frac { -3 }{ -5 } \) = \(\frac { 3 }{ 5 } \)

Slope of the second line (m_{2}) = \(\frac { -5 }{ k } \)

Since the two lines are perpendicular.

m_{1} × m_{2} = -1

\(\frac { 3 }{ 5 } \) × \(\frac { -5 }{ k } \) = -1 ⇒ \(\frac { -3 }{ k } \) = -1

-k = -3 ⇒ The value of k = 3

Question 12.

If x – y = 3 and x + 2y = 6 are the diameters of a circle then the centre is at the point ………..

(1) (0, 0)

(2) (1, 2)

(3) (1, -1)

(4) (4, 1)

Answer:

(4) (4, 1)

Hint:

Centre of the circle is the intersection of the two diameters.

Centre of the circle is (4, 1)

Question 13.

The line 4x + 3y – 12 = 0 meets the x-axis at the point ……….

(1) (4, 0)

(2) (3, 0)

(3) (-3, 0)

Answer:

(2) (3,0)

Hint:

4x + 3y – 12 = 0 meet the x-axis the value of y = 0

4x- 12 = 0 ⇒ 4x = 12

x = \(\frac { 12 }{ 4 } \) = 3 ⇒ The point is (3, 0)

Question 14.

The equation of a straight line passing through the point (2, -7) and parallel to x-axis is ……………….

(1) x = 2

(2) x = -7

(3) y = -7

(4) y = 2

Answer:

(3) y = -7

Hint:

Equation of a line parallel to x-axis is y = -7

Question 15.

The equation of a straight line having slope 3 and y intercept – 4 is ………………

(1) 3x – y – 4 = 0

(2) 3x + y – 4 = 0

(3) 3x – y + 4 = 0

(4) 3x – y + 4 = 0

Answer:

(1) 3x – y – 4 = 0

Hint. The equation of a line is y = mx + c

y = 3 (x) + (-4) ⇒ y = 3x – 4

3x – y – 4 = 0

II. Answer the following questions:

Question 1.

If the points (3, – 4) (1, 6) and (- 2, 3) are the vertices of a triangle, find its area.

Answer:

Let the vertices A (3, – 4), B (1, 6) and C (- 2, 3)

Area of ∆ ABC = \(\frac { 1 }{ 2 } \) [x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1}, – (x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3})]

Area of a ∆ = 18 sq. units

Question 2.

If the area of the triangle formed by the points (1,2) (2,3) and (a, 4) is 8 sq. units, find a.

Answer:

Area of a triangle = 8 sq. units.

\(\frac { 1 }{ 2 } \) [x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} – (x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3})] = 8.

\(\frac { 1 }{ 2 } \) [3 + 8 + 2a – (4 + 3a + 4)] = 8

11 + 2a – 8 – 3a= 16 ⇒ – a + 3 = 16

– a = 16 – 3 ⇒ a = -13

The value of a = -13

Question 3.

If the points A (2, 5), B (4, 6) and C (8, a) are collinear find the value of “a” using slope concept.

Answer:

Since the three points are collineal

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Slope of AB = Slope of BC

\(\frac { 6-5 }{ 4-2 } \) = \(\frac { a-6 }{ 8-4 } \) ⇒ \(\frac { 1 }{ 2 } \) = \(\frac { a-6 }{ 4 } \) ⇒ 2a – 12 = 4 ⇒ 2a = 16

a = \(\frac { 16 }{ 2 } \) = 8 ⇒ The value of a = 8

Question 4.

If the points (x,y) is collinear with the points (a, 0) and (0, b) then prove that \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1

Answer:

Let A (x, y), B (a, 0), C(0, b)

Since the three points are collinear

Slope of AB = Slope of BC

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

\(\frac { 0-y }{ a-x } \) = \(\frac { b-0 }{ 0-a } \)

\(\frac { -y }{ a-x } \) = \(\frac { b }{ -a } \)

ay = b (a – x)

ay = ba – bx

ay + bx = ab

Divided by ab

\(\frac { ay }{ ab } \) + \(\frac { bx }{ ab } \) = \(\frac { ab }{ ab } \)

\(\frac { y }{ b } \) + \(\frac { x }{ a } \) = 1 ⇒ \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1

Question 5.

A straight line passes through (1, 2) and has the equation y – 2x – k = 0. Find k.

Answer:

The given line is y – 2x – k = 0

It passes through (1,2)

(2) -2 (1) -k = 0 ⇒ 2 – 2 – k = 0

0 – k = 0 ⇒ k = 0

The value of k = 0

Question 6.

If a line passes through the mid point of AB where A is (3, 0) and B is (5, 4) and makes an angle 60° with x – axis find its equation.

Answer:

Slope of a line = tan 60°

= \(\sqrt { 3 }\)

Equation of a line is y – y_{1} = m (x – x_{1})

y – 2 = \(\sqrt { 3 }\) (x – 4)

y – 2 = \(\sqrt { 3 }\) x – 4 \(\sqrt { 3 }\)

\(\sqrt { 3x }\) – y + 2 – 4\(\sqrt { 3 }\) = 0

Question 7.

Find the equation of the line through (3, 2) and perpendicular to the line joining (4, 5) and (1,2)

Answer:

Slope of a line = \(\frac { 2-5 }{ 1-4 } \) ⇒ \(\frac { -3 }{ -3 } \) = 1

Slope of the line perpendicular to it is – 1

Equation of the line joining -1 and (3, 2) is

y – y_{1} = m (x – x_{1}) ⇒ y – 2 = -1(x – 3)

y – 2 = -x + 3 ⇒ x + y – 5 = 0

Question 8.

P and Q trisect the line segment joining the points (2, 1) and (5, – 8). If the point P lies on 2x – y + k = 0, then find the value of k.

Answer:

A line divides internally in the ratio 1 : 2

A line divide internally in the ratio l : m

The point P = (\(\frac { 5+4 }{ 3 } \),\(\frac { -8+2 }{ 3 } \))

= (\(\frac { 9 }{ 3 } \),\(\frac { -6 }{ 3 } \)) = (3, -2)

The given line 2x – y + k = 0 passes through the point (3,-2)

2 (3) – (- 2) + k = 0

6 + 2 + k = 0

8 + k = 0

k = – 8

The value of k = – 8

Question 9.

The line 4x + 3y – 12 = 0 intersect the X, Y – axis at A and B respectively. Fine the area of ∆AOB.

Answer:

The equation of the line AB is 4x + 3y – 12 = 0

4x + 3y = 12

\(\frac { 4x }{ 12 } \) + \(\frac { 3y }{ 12 } \) = 1 ⇒ \(\frac { x }{ 3 } \) + \(\frac { y }{ 4 } \) = 1

The point A is (3, 0) (it intersect the X – axis)

and B is (0, 4) (it intersect the Y – axis)

Area of ∆ AOB = \(\frac { 1 }{ 2 } \) [x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} (x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3})]

Question 10.

Find the equation of the line passing through (4, 5) and making equal intercept in the axes.

Answer:

Let the equal intercept on the axes be a, a.

Equation of the line is \(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1 (Given equal intercepts)

The line passes through (4, 5)

\(\frac { 4 }{ a } \) + \(\frac { 5 }{ a } \) = 1 ⇒ \(\frac { 9 }{ a } \) = 1 ⇒ a = 9

The equation of the line is \(\frac { x }{ 9 } \) + \(\frac { y }{ 9 } \) = 1

Multiply by 9

x + y – 9 = 0

Question 11.

Find the equation of the line passing through (2, – 1) and whose intercepts on the axes are equal in magnitude but opposite in sign.

Answer:

Let the x – intercept be “a” and y intercept be = “-a”

The equation of the line is

\(\frac { x }{ a } \) + \(\frac { y }{ -a } \) = 1 (y – intercept is – a)

\(\frac { x }{ a } \) – \(\frac { y }{ a } \) = 1

It passes through (2, -1)

\(\frac { 2 }{ a } \) – \(\frac { (-1) }{ a } \) = 1

\(\frac { 2 }{ a } \) + \(\frac { 1 }{ a } \) = 1 ⇒ \(\frac { 3 }{ a } \) = 1

a = 3

The equation of the line is

\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1

\(\frac { x }{ 3 } \) + \(\frac { y }{ -3 } \) = 1 ⇒ \(\frac { x }{ 3 } \) – \(\frac { y }{ 3 } \) = 1

x – y = 3

The equation is x – y – 3 = 0

Question 12.

The straight line cuts the coordinate axes at A and B. If the mid point of AB is (3,2) then find the equation of AB.

Answer:

Let the point A be (a, 0) and B be (0, b)

The point A (6, 0) and B (0, 4)

Equation of the line AB is

III. Answer the following questions

Question 1.

If the coordinates of two points A and B are (3, 4) and (5, – 2) respectively. Find the ‘ coordinates of any point “c”, if AC = BC and Area of triangle ABC = 10 sq. units.

Answer:

Let the coordinates C be (a, 6) then AC = BC

AC^{2} = BC^{2}

(a – 3)^{2} + (b – 4)^{2} = (a – 5)^{2} + (b + 2)^{2}

a^{2} + 9 – 6a + b^{2} + 16 – 8b = a^{2} + 25 – 10a + b^{2} + 4 – 4b

a^{2} + b^{2} + 25 – 6a – 86 = a^{2} + b^{2} + 29 – 10a + 4b

25 – 6a – 8b = 29 – 10a + 46

4a – 12b = 4 ⇒ a – 3b = 1 ………… (1)

Area of ∆ ABC = 10 sq. units

\(\frac { 1 }{ 2 } \) [x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} – (x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3})] = 10

-6 + 5b + 4a – (20 – 2a + 3b) = 20

-6 + 5b + 4a – 20 + 2a – 3b = 20

6a + 2b – 26 = 20 ⇒ 6a + 2b = 46

Substitute the value of a = 7 in (2)

3 (7) + b = 23 ⇒ b = 23 – 21 = 2

The coordinate C is (7, 2)

Question 2.

The four vertices of a Quadrilateral are (1,2) (- 5,6) (7, – 4) and (k, – 2) taken in order. If the area of the Quadrilateral is 9 sq. units, find the value of k.

Answer:

Let A (1, 2) B (- 5, 6) C (7, – 4) and D (k, – 2)

Area of the

Quadrilateral ABCD = \(\frac { 1 }{ 2 } \)[(x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{4} + x_{4}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{3} + x_{1}y_{4})]

Area of the Quadrilateral ABCD = 3k – 9

Given area of a Quadrilateral is 9 sq. units.

3k – 9 = 9 ⇒ 3k = 18 ⇒ k = \(\frac { 18 }{ 3 } \) = 6

The value of k = 6

Question 3.

Find the area of a triangles whose three sides are having the equations x + y = 2, x – y = 0 and x + 2y – 6 = 0.

Answer:

Find the three vertices of the triangles by solving their equation.

To find vertices A

Substitute the value of y = 4 in (1)

x + 4 = 2 ⇒ x = 2 – 4 = -2

The vertices A is (- 2, 4)

To find vertices B

Substitute the value of x = 1 in (1)

1 + y = 2 ⇒ y = 2 – 1 = 1

The vertices B is (1, 1)

To find vertices C

y = \(\frac { 6 }{ 3 } \) = 2

Substitute the value y = 2 in (3)

x – 2 = 0 ⇒ x = 2

The vertices C is (2, 2)

Area of the ∆ BC = 3 sq. units

Question 4.

Verify the Median of a triangle divides into two triangles of equal areas whose vertices are A (4, – 6), B (3, – 2) and C (5, 2)

Answer:

Let D be the mid point of AC .

Mid point of AC = (\(\frac { 5+4 }{ 2 } \),\(\frac { 2-6 }{ 2 } \)) = (\(\frac { 9 }{ 2 } \),-2)

Area of the triangle = \(\frac { 1 }{ 2 } \) [x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} – (x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3})]

Area of ∆ ADB = Area of ∆ BDC

A median divides the triangle of equal areas.

Question 5.

Find the area of the ∆ ABC with A (1, – 4) and the mid points of sides through A being (2,-1) and (0,-1)

Answer:

Let the coordinates of B and C are (a, b) and (c, d) respectively.

Sides through A are AB and AC

Mid point of AB = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))

(2, -1) = (\(\frac { 1+a }{ 2 } \),\(\frac { -4+b }{ 2 } \))

\(\frac { 1+a }{ 2 } \) = 2

1 + a = 4

a = 4 – 1

= 3

The point B is (3,2)

\(\frac { -4+b }{ 2 } \) = -1

-4 + b = -2

b = -2 + 4

= 2

Mid point of AC = (\(\frac { 1+c }{ 2 } \),\(\frac { -4+d }{ 2 } \))

(0,-1) = (\(\frac { 1+c }{ 2 } \),\(\frac { -4+d }{ 2 } \))

\(\frac { 1+c }{ 2 } \) = 0

1 + c = 0

c = 0 – 1

= – 1

The point C is (-1,2)

\(\frac { -4+d }{ 2 } \) = -1

– 4 + d = -2

d = – 2 + 4

= 2

Thus the coordinates of the vertices of ∆ ABC are A (1, – 4) B (3, 2) and C (- 1, 2)

Area of ∆ ABC = \(\frac { 1 }{ 2 } \) [x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} – (x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3})]

Area of ∆ ABC = 12 sq. units

Question 6.

Find the equation of the straight lines passing through (- 3, 10) whose sum of the intercepts is 8.

Answer:

Let the “x” intercept be “a” and y intercept be “b”

Sum of the intercepts = 8

a + b = 8 ⇒ b = 8 – a

Equation of a line is \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1 ⇒ \(\frac { x }{ a } \) + \(\frac { y }{ 8-a } \) = 1

It passes through (-3,10)

\(\frac { -3 }{ a } \) + \(\frac { 10 }{ 8-a } \) = 1

\(\frac { -3(8-a)+10a }{ a(8-a) } \) = 1

-24 + 3a + 10a = 8a – a^{2}

-24 + 13a = 8a – a^{2}

a^{2} + 5a – 24 = 0 ⇒ (a + 8) (a – 3) = 0

a + 8 = 0 (or) a – 3 = 0 ⇒ a = -8 (or) a = 3

The equation of a line is a

a = -8

\(\frac { x }{ -8 } \) + \(\frac { y }{ 8+8 } \) = 1

\(\frac { x }{ -8 } \) + \(\frac { y }{ 16 } \) = 1

-2x + y = 16

2x – y + 16 = 0

a = 3

\(\frac { x }{ 3 } \) + \(\frac { y }{ 5 } \) = 1

5x + 3y = 15

5x + 3y – 15 = 0

The equation of the lines are 2x – y + 16 = 0 (or) 5x + 3y – 15 = 0.

Question 7.

If (5, – 3), (- 5, 3), (6, 6) are the mid points of the sides of a triangle, find the equation of the sides.

Answer:

Since D, E, F are the mid points of ∆ ABC

EF || AB, FD || BC and DE || AC

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Slope of EF = \(\frac { 6-3 }{ 6+5 } \) = \(\frac { 3 }{ 11 } \)

Since EF || AB; Slope of AB = \(\frac { 3 }{ 11 } \)

Equation of AB is

y – y_{1} = m (x – x_{1})

y + 3 = \(\frac { 3 }{ 11 } \) (x – 5)

3x – 15 = 11y + 33

3x – 11y – 15 – 33 = 0

3x – 11y – 48 = 0

Slope of DE = Slope of AC

Slope of DE = \(\frac { 3+3 }{ -5-5 } \) = \(\frac { 6 }{ -10 } \) = –\(\frac { 6 }{ 10 } \) = –\(\frac { 3 }{ 5 } \)

Slope of AC = – \(\frac { 3 }{ 5 } \)

Equation of AC is

y – y_{1} = m (x – x_{1})

y – 6 = – \(\frac { 3 }{ 5 } \) (x – 6) ⇒ 5y – 30 = -3x + 18

3x + 5y – 30 – 18 = 0 ⇒ 3x + 5y – 48 = 0

Slope of DF = Slope of BC

Slope of DF = \(\frac { 6+3 }{ 6-5 } \) = \(\frac { 9 }{ 1 } \) = 9

Slope of BC = 9

Equation of the line BC is

y – y_{1} = m(x – x_{1})

y – 3 = 9 (x + 5) ⇒ 9x + 45 = y – 3

9x – y + 45 + 3 = 0 ⇒ 9x – y + 48 = 0

Equation of the sides are

3x – 11y – 48 = 0 ; 9x – y + 48 = 0 and 3x + 5y – 48 = 0

Question 8.

Find the equation of the straight line passing through the point of intersection of the lines 5x – 8y + 23 = 0 and 7x + 6y – 71 = 0 and is perpendicular to the line joining the points (5,1) and (-2, 2)

Answer:

Substitute the value of x in (1)

5 (5) – 8y = – 23 ⇒ 25 – 8y = – 23

-8y = – 23 – 25 ⇒ -8y = – 48

y = \(\frac { 48 }{ 8 } \) = 6

The point of intersection is (5,6)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

Slope of the line joining the points (5,1) and (-2,2) = \(\frac { 2-1 }{ -2-5 } \)

= \(\frac { 1 }{ -7 } \) = – \(\frac { 1 }{ 7 } \)

Slope of the perpendicular line is = 7

Equation of a line is

y – y_{1} = m(x – x_{1}) ⇒ y – 6 = 7 (x – 5)

y – 6 = 7x – 35 ⇒ -7x + y – 6 + 35 = 0

7x – y – 29 = 0

Question 9.

Find the equation of the line passing through the point of intersection of 4x – y – 3 = 0 and x + y – 2 = 0 and perpendicular to 2x – 5y + 3 = 0.

Answer:

x = \(\frac { 5 }{ 5 } \) = 1

Substitute the value of x = 1 in (2)

1 + y = 2

y = 2 – 1 = 1

The point of intersection is (1, 1)

Any line perpendicular to 2x – 5y + 3 = 0 is

5x + 2y + k = 0

It passes through (1,1)

5(1) + 2(1) + k = 0 ⇒ 5 + 2 + k = 0

7 + k = 0 ⇒ k = -7

The line is 5x + 2y – 7 = 0

Question 10.

Find the equation of the line through the point of intersection of the lines 2x + y – 5 = 0 and x + y – 3 = 0 and bisecting the line segment joining the points (3, – 2) and (- 5, 6).

Answer:

x = 2

Substitute the value of x = 2 in (2)

2 + y = 3

y = 3 – 2 = 1

The point of intersection is (2, 1)

Mid point of the line joining the points (3,-2) and (-5,6)

Mid point of the line

Equation of the line joining the points (2, 1) and (-1,2) is

x – 2 = -3 (y – 1)

x – 2 = -3y + 3

x + 3y – 5 = 0

The equation of the line is x + 3y – 5 = 0