Coordinates of mid point of line joining points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ = $$(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2})$$

Let coordinates of Point A = $$(x , y)$$

P = (3,1) and B = (5,-4)

=> $$3 = \frac{x+5}{2}$$

=> $$x = 6-5 = 1$$

Similarly, $$1 = \frac{y -4}{2}$$

=> $$y = 2+4 = 6$$

Thus, coordinates of Point A = (1,6)

=> Ans - (C)

$$(x,y)=(\frac{mx_{2}+nx_{1}}{m+n},\frac{my_{2}+ny_{1}}{m+n})$$

$$(x,y)=(\frac{4(7/2)+1(6)}{4+1},\frac{4(6)+1(1)}{4+1})$$

$$(x,y)=(4,5)$$

So the answer is option B.

Let line $$l$$ perpendicularly bisects line joining  A(2,-5) and B(0,7) at C, thus C is the mid point of AB.

=> Coordinates of C = $$(\frac{2 + 0}{2} , \frac{-5 + 7}{2})$$

= $$(\frac{2}{2} , \frac{2}{2}) = (1,1)$$

Now, slope of AB = $$\frac{y_2 - y_1}{x_2 - x_1} = \frac{(7 + 5)}{(0 - 2)}$$

= $$\frac{12}{-2} = -6$$

Let slope of line $$l = m$$

Product of slopes of two perpendicular lines = -1

=> $$m \times -6 = -1$$

=> $$m = \frac{1}{6}$$

Equation of a line passing through point $$(x_1,y_1)$$ and having slope $$m$$ is $$(y - y_1) = m(x - x_1)$$

$$\therefore$$ Equation of line $$l$$

=> $$(y - 1) = \frac{1}{6}(x - 1)$$

=> $$6y - 6 = x - 1$$

=> $$x - 6y = 1 - 6 = -5$$

=> Ans - (C)

Coordinates of centroid of triangle with vertices $$(x_1 , y_1)$$ , $$(x_2 , y_2)$$ and $$(x_3 , y_3)$$ is $$(\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3})$$

Let coordinates of vertex C = $$(x , y)$$

Vertex A(6,-4) and Vertex B(-2,2) and Centroid = (-1,-2)

=> $$-1 = \frac{-2 + 6 + x}{3}$$

=> $$x + 4 = -1 \times 3 = -3$$

=> $$x = -3 - 4 = -7$$

Similarly, => $$-2 = \frac{-4 + 2 + y}{3}$$

=> $$y - 2 = -2 \times 3 = -6$$

=> $$y = -6 + 2 = -4$$

$$\therefore$$ Coordinates of vertex C = (-7,-4)

=> Ans - (A)

Using section formula, the coordinates of point that divides line joining A = $$(x_1 , y_1)$$ and B = $$(x_2 , y_2)$$ in the ratio a : b

= $$(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$$

Coordinates of A(0,4) and B(-5,9). Let coordinates of P = (x,y) which divides AB in ratio = 2 : 3

=> $$x = \frac{(2 \times -5) + (3 \times 0)}{2 + 3}$$

=> $$5x = -10$$

=> $$x = \frac{-10}{5} = -2$$

Similarly, $$y = \frac{(2 \times 9) + (3 \times 4)}{2 + 3}$$

=> $$5y = 18 + 12 = 30$$

=> $$y = \frac{30}{5} = 6$$

=> Point P = (-2,6)

Slope of AB = $$\frac{9 - 4}{-5 - 0} = \frac{5}{-5} = -1$$

Let slope of line perpendicular to AB = $$m$$

Also, product of slopes of two perpendicular lines is -1

=> $$m \times -1 = -1$$

=> $$m = 1$$

Equation of lines having slope $$m$$ and passing through point P(-2,6) is

=> $$(y - 6) = 1(x + 2)$$

=> $$y - 6 = x + 2$$

=> $$x - y = -8$$

=> Ans - (B)

Using section formula, the coordinates of point that divides line joining A = $$(x_1 , y_1)$$ and B = $$(x_2 , y_2)$$ in the ratio a : b

= $$(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$$

Let the ratio in which the segment joining (12,1) and (3,4) divided by the y-axis = $$k$$ : $$1$$

Since, the line segment is divided by y-axis, thus x coordinate of the point will be zero, let the point of intersection = $$(0,y)$$

Now, point P (0,y) divides (12,1) and (3,4) in ratio = k : 1

=> $$0 = \frac{(3 \times k) + (12 \times 1)}{k + 1}$$

=> $$3k + 12 = 0$$

=> $$k = \frac{-12}{3} = -4$$

$$\therefore$$ Line segment joining (12,­1) and (­3,4) is divided by the Y ­axis in the ratio = 4 : 1 externally

=> Ans - (A)

Reflection of point $$(x,y)$$ in the line $$x=a$$ is = $$(-x+2a,y)$$

and in line $$y=b$$ is = $$(x,-y+2b)$$

Thus, reflection of the point (5, 2) in the line x = -3

= $$[(-5+2\times-3),2]$$

= $$(-11,2)$$

=> Ans - (A)

centroid = $$(\frac{x_{1}+x_{2}+x_{3}}{3} , \frac{y_{1}+y_{2}+y_{3}}{3})$$ = $$(\frac{1+4-2}{3} , \frac{-5+0+2}{3})$$ = $$(1 , -1)$$

so the answer is option A.

slope = m = -a/b = -5/3, c = -3

eq of perpendicular line is

y = (-1/m)x + c

y = (3/5)x - 3

5y = 3x - 15

3x - 5y = 15

so the answer is option A.

y = mx + c

y = -x/3 + 6

3y = -x + 18

x + 3y = 18

so the answer is option A.

Slope of line passing through points (2,1) and (6,­3)

= $$\frac{3 - 1}{6 - 2} = \frac{2}{4} = \frac{1}{2}$$

Slope of two parallel lines is always equal.

=> Slope of the line parallel to the line having slope 1/2 = $$\frac{1}{2}$$

=> Ans - (A)

Slope of line passing through points (4,-2) and (-3,5)

= $$\frac{5 + 2}{-3 - 4} = \frac{7}{-7} = -1$$

Slope of two parallel lines is always equal.

=> Slope of the line parallel to the line having slope -1 = $$-1$$

=> Ans - (D)

Equation : 5x + 4y = 2

(A) : (2,3)

=> $$5(2) + 4(3) = 22 \neq 2$$

(B) : (-2,-3)

=> $$5(-2) + 4(-3) = -22 \neq 2$$ 

(C) : (-2,3)

=> $$5(-2) + 4(3) = 2$$

(D) : (2,-3)

=> $$5(2) + 4(-3) = -2 \neq 2$$

=> Ans - (C)

Mid-point = $$(\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2})$$  $$\Rightarrow \frac{-5+x}{2}=-2 \Rightarrow x = -4 + 5 = 1$$

so the answer is option A.

(diagram)

From the diagram, the reflection of the point (2, 3) in the line y = 4 is (2 , 5)

so the answer is option A.

Slope of line 3x - 2y = 6 = $$-\frac{3}{-2} = \frac{3}{2}$$

Product of slopes of two perpendicular lines = -1

Let slope of line passing through origin = $$m$$

=> $$m \times \frac{3}{2} = -1$$

=> $$m = \frac{-2}{3}$$

Equation of line passing through origin and having slope m is $$y = mx$$    (Since y intercept is zero)

=> $$y = \frac{-2}{3} x$$

=> $$3y = -2x$$ => $$2x + 3y = 0$$ 

Solving the above equations, we get the intersection point M = $$(\frac{18}{13} , \frac{-12}{13})$$

=> Ans - (B)

Coordinates of mid point of line joining A $$(x_1,y_1)$$ and B $$(x_2,y_2)$$ = $$(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2})$$

Coordinates of E $$(-4 , a)$$ and F $$(2 , 0)$$

Q $$(b , -1)$$ is the mid point of EF

=> $$b = \frac{-4 + 2}{2}$$

=> $$b = \frac{-2}{2} = -1$$

Similarly, $$-1 = \frac{a + 0}{2}$$

=> $$a = -1 \times 2 = -2$$

=> Ans - (B)

Equation of line passing thorough points $$(x_1 , y_1)$$ and $$(x_2 , y_2)$$ = $$(y - y_1) = \frac{(y_2 - y_1)}{(x_2 - x_1)} (x - x_1)$$

x-intercept of 4 units implies that the line passes through (4,0)

Points given = (-1,3) and (4,0)

=> Equation of line = $$(y - 3) = \frac{(0 - 3)}{(4 + 1)} (x + 1)$$

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

=> $$5y - 15 = -3x - 3$$

=> $$3x + 5y = 12$$

=> Ans - (B)

Using section formula, the coordinates of point that divides line joining A = $$(x_1 , y_1)$$ and B = $$(x_2 , y_2)$$ in the ratio a : b

= $$(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$$

Now, point P (-4,1) divides (x,0) and (0,y) in ratio = 3 : 2

=> $$-4 = \frac{(3 \times 0) + (2 \times x)}{3 + 2}$$

=> $$x = \frac{-4 \times 5}{2} = -10$$

Similarly, $$1 = \frac{(3 \times y) + (2 \times 0)}{3 + 2}$$

=> $$3y = 5$$, => $$y = \frac{5}{3}$$

=> Ans - (A)

(diagram)

From the diagram, the reflection of the point (4, -3) in the line y = -2 is (4 , -1)

so the answer is option D.

Slope of line passing through $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$\frac{y_2 - y_1}{x_2 - x_1}$$

=> Slope of line passing through (1,2) and (3,0) = $$\frac{0 - 2}{3 - 1} = \frac{-2}{2} = -1$$

Slope of line passing through (4,3) and (y,0) = $$\frac{0 - 3}{y - 4} = \frac{-3}{(y - 4)}$$

Also, slopes of parallel lines are equal.

=> $$\frac{-3}{y - 4} = -1$$

=> $$y - 4 = 3$$

=> $$y = 3 + 4 = 7$$

=> Ans - (B)

Diagonals of a rhombus bisect each other perpendicularly. Thus, O is the mid point of QS and RP.

Coordinates of midpoint of line joining A = $$(x_1 , y_1)$$ and B = $$(x_2 , y_2)$$ is $$(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2})$$

=> $$O = (\frac{3 - 7}{2} , \frac{1 + 5}{2}) = (-2 , 3)$$

Slope of PR = $$\frac{1 - 5}{3 + 7} = \frac{-2}{5}$$

Product of slopes of two perpendicular lines is -1

=> Slope of QS = $$m \times \frac{-2}{5} = -1$$

=> $$m = \frac{5}{2}$$

$$\therefore$$ Equation of line QS with slope $$m = \frac{5}{2}$$ and passing through point O $$(x_1,y_1) = (-2,3)$$

= $$(y - y_1) = m (x - x_1)$$

= $$(y - 3) = \frac{5}{2} (x + 2)$$

= $$2y - 6 = 5x + 10$$

= $$5x - 2y = -16$$

The line $$3x + y = -6$$ will intercept the x-axis at $$y = 0$$

Thus, substituting value of y in above equation

=> $$3x + 0 = -6$$

=> $$x = \frac{-6}{3} = -2$$

Thus, the line will intercept x axis at (-2,0)

=> Ans - (B)

$$\sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}$$ = $$13$$

$$(-5-7)^{2}+(k-2)^{2}$$ = $$169$$

$$144+(k-2)^{2}$$ = $$169$$

$$(k-2)^{2}$$ = $$169-144$$

$$k-2 = 5$$

$$ k = 7$$

So the answer is option B.

slope m = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-4+1}{4-5}=3$$

slope of the parallel line = m = 3.

So the answer is option C.

(diagram)

from the diagram answer is (-7 , -2)

So the answer is option A.

Slope, $$m = \frac{-2}{5}$$ and y-intercept, $$c = 6$$

Equation of line = $$y = mx + c$$

=> $$y = \frac{-2}{5} x + 6$$

=> $$y = \frac{-2x + 30}{5}$$

=> $$5y = -2x + 30$$

=> $$2x + 5y = 30$$

=> Ans - (B)

Slope of line 2x - ­3y = 11 is $$\frac{-2}{-3}$$

= $$\frac{2}{3}$$

Slope of line 3x + ky = ­4 is $$\frac{-3}{k}$$

Also, product of slopes of two perpendicular lines is -1

=> $$\frac{2}{3} \times \frac{-3}{k} = -1$$

=> $$\frac{-2}{k} = -1$$

=> $$k = -2 \times -1 = 2$$

=> Ans - (D)

Reflection of point (x,y) in line x=a is (-x+2a,y)

Now, Reflection of point (-2,5) in line x=-1

= [2+2(-1),5]=(0,5)

=> Ans - (B)

Using section formula, the coordinates of point that divides line joining A = $$(x_1 , y_1)$$ and B = $$(x_2 , y_2)$$ in the ratio a : b

= $$(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$$

Let the ratio in which the segment joining (2,3) and (-2,1) divided by the y-axis = $$k$$ : $$1$$

Since, the line segment is divided by y-axis, thus x coordinate of the point will be zero, let the point of intersection = $$(0,y)$$

Now, point P (0,y) divides (2,3) and (-2,1) in ratio = k : 1

=> $$0 = \frac{(-2 \times k) + (2 \times 1)}{k + 1}$$

=> $$-2k + 2 = 0$$

=> $$k = \frac{2}{2}=1$$

$$\therefore$$ Required ratio = 1 : 1

=> Ans - (B)

Q(a,b) after reflection in the y-axis = (-a,b)

Reflection of point (-a,b) in the x-axis is (-a,-b)

According to ques,

=> $$(-a,-b) = (-6,2)$$

=> $$-a = -6$$ and $$-b = 2$$

$$\therefore$$ Coordinates of Point Q = (6,-2)

=> Ans - (C)

Coordinates of mid point of line joining points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ = $$(\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2})$$

Let coordinates of mid point $$(x , y)$$

C = (3,-5) and D = (-7,3)

=> $$x = \frac{3 - 7}{2}$$

=> $$x = \frac{-4}{2} = -2$$

Similarly, $$y = \frac{-5 + 3}{2}$$

=> $$y = \frac{-2}{2} = -1$$

Thus, coordinates of mid point (-2,-1)

Slope of line $$ax+by=c$$ is = $$\frac{-a}{b}$$

=> Slope of 2x - 5y = 12 is = $$\frac{-2}{-5}$$

= $$\frac{2}{5}$$

=> Ans - (A)

Let the line be $$l$$ which has x-intercept -5, => $$l$$ passes through (-5,0)

Slope of line $$ax + by + c = 0$$ is $$\frac{-a}{b}$$

=> Slope of line $$y = 2x + 3$$ => $$2x - y + 3 = 0$$

=> Slope = $$\frac{-2}{-1} = 2$$

Product of slopes of two perpendicular lines = -1

Let slope of line $$l$$ = $$m$$

=> $$m \times 2 = -1$$

=> $$m = \frac{-1}{2}$$

Now, equation of line having slope $$m$$ and passing through $$(x_1 , y_1)$$ is $$(y - y_1) = m(x - x_1)$$

=> $$(y - 0) = \frac{-1}{2} (x + 5)$$

=> $$2y = -x - 5$$

=> $$x + 2y = -5$$

Let line $$l$$ perpendicularly bisects line joining  A(5,-3) and B(0,2) at C, thus C is the mid point of AB.

=> Coordinates of C = $$(\frac{5 + 0}{2} , \frac{-3 + 2}{2})$$

= $$(\frac{5}{2} , \frac{-1}{2})$$

Now, slope of AB = $$\frac{y_2 - y_1}{x_2 - x_1} = \frac{(2 + 3)}{(0 - 5)}$$

= $$\frac{5}{-5} = -1$$

Let slope of line $$l = m$$

Product of slopes of two perpendicular lines = -1

=> $$m \times -1 = -1$$

=> $$m = 1$$

Equation of a line passing through point $$(x_1,y_1)$$ and having slope $$m$$ is $$(y - y_1) = m(x - x_1)$$

$$\therefore$$ Equation of line $$l$$

=> $$(y + \frac{1}{2}) = 1(x - \frac{5}{2})$$

=> $$x - y = \frac{5}{2} + \frac{1}{2} = \frac{6}{2}$$

=> $$x - y = 3$$

=> Ans - (D)

Slope of the line passing through the points (-5 , 1) and (-2 , 0) = $$\frac{0-1}{-2+5}=\frac{-1}{3}$$

Let slope of line perpendicular to it = $$m$$

Also, product of slope of two perpendicular lines = -1

=> $$m\times \frac{-1}{3}=-1$$

=> $$m=(-1)\times(-3)=3$$

=> Ans - (B)

Slope of line 2x + 3y = 6 is $$(-\frac{2}{3})$$

Product of slopes of two perpendicular lines = -1

Let slope of line passing through origin = $$m$$

=> $$m \times \frac{-2}{3} = -1$$

=> $$m = \frac{3}{2}$$

Equation of line passing through origin and having slope m is $$y = mx$$    (Since y intercept is zero)

=> $$y = \frac{3}{2} x$$

=> $$3x = 2y$$ 

Solving the above equations, we get the intersection point M = $$(\frac{12}{13} , \frac{18}{13})$$

=> Ans - (A)

A line which intercept x-axis at $$(-3,0)$$ and y-axis at $$(0,6)$$

=> Slope, $$m = \frac{6-0}{0+3} = \frac{6}{3} = 2$$

Equation of line passing thorough points $$(x_1 , y_1)$$ and slope $$m$$ = $$(y - y_1) = m (x - x_1)$$

=> Equation of line = $$(y - 0) = 2 (x + 3)$$

=> $$y = 2x + 6$$

=> Ans - (D)

P(a,b) after reflection at the origin = (-a,-b)

Reflection of point (-a,-b) in the y-axis is (a,-b)

According to ques,

=> $$(a,-b) = (4,3)$$

=> $$a = 4$$ and $$b = -3$$

$$\therefore$$ Coordinates of Point P = (4,-3)

=> Ans - (A)

Slope of line making an angle of 45° with the positive x­-axis = $$tan(45$$°$$)$$

=> Slope, $$m = 1$$

y-intercept, $$c = \frac{3}{4}$$

Equation of line having slope $$m$$ and y intercept $$c$$ is : $$y = mx+c$$

=> $$y = x + \frac{3}{4}$$

=> $$4y = 4x+3$$

=> $$4x-4y=-3$$

=> Ans - (B)

Reflection of point (x,y) in line y=a is (x,-y+2a)

Now, Reflection of point (5,-1) in line y=2

= [5,1+2(2)]=(5,5)

=> Ans - (C)

Slope of line having equation : $$ax + by + c = 0$$ is $$\frac{-a}{b}$$

=> Slope of line $$20x + 5y = 3$$ is $$\frac{-20}{5} = -4$$

Slope line passing through (-2,5) and (6,b) = $$\frac{b - 5}{6 + 2} = \frac{(b - 5)}{8}$$

Also, product of slopes of two perpendicular lines is -1

=> $$\frac{(b - 5)}{8} \times -4 = -1$$

=> $$b - 5 = \frac{8}{4} = 2$$

=> $$b = 2 + 5 = 7$$

=> Ans - (C)

slope = $$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3+5}{-5-x}=\frac{8}{-5-x}=\frac{-4}{3}$$

$$\rightarrow 24 = 20 + 4x$$

$$\rightarrow x = 1$$

so the answer is option D.

When a line intercepts y-axis at a point, then x-coordinate of that point is 0.

Let the line intercepts y-axis at $$(0,y)$$

Equation of line = $$4x - 3y = -6$$

Putting $$x = 0$$ in above equation, we get :

=> $$(4 \times 0) - 3y = -6$$

=> $$3y = 6$$

=> $$y = \frac{6}{3} = 2$$

$$\therefore$$ The line 4x - 3y = -6 will intercept the y-axis at = (0,2)

=> Ans - (A)

Using section formula, the coordinates of point that divides line joining A = $$(x_1 , y_1)$$ and B = $$(x_2 , y_2)$$ in the ratio a : b

= $$(\frac{a x_2 + b x_1}{a + b} , \frac{a y_2 + b y_1}{a + b})$$

Let the ratio in which the segment joining (-1,-12) and (3,4) divided by the x-axis = $$k$$ : $$1$$

Since, the line segment is divided by x-axis, thus y coordinate of the point will be zero, let the point of intersection = $$(x,0)$$

Now, point P (x,0) divides (-1,-12) and (3,4) in ratio = k : 1

=> $$0 = \frac{(4 \times k) + (-12 \times 1)}{k + 1}$$

=> $$4k - 12 = 0$$

=> $$k = \frac{12}{4} = 3$$

$$\therefore$$ Required ratio = 3 : 1

=> Ans - (C)

Slope of line : $$ax-4y=-6$$ = $$\frac{a}{4}$$

According to ques,

=> $$\frac{a}{4}=\frac{-3}{2}$$

=> $$a=\frac{-3}{2} \times 4=-6$$

=> Ans - (C)

Reflection of point (x,y) in line x=a is (-x+2a,y)

Now, Reflection of point (-3,1) in line x=-2

= [3+2(-2),1]=(-1,1)

=> Ans - (A)

Let line $$l$$ perpendicularly bisects line joining  A(2,-6) and B(4,0) at C, thus C is the mid point of AB.

=> Coordinates of C = $$(\frac{2 + 4}{2} , \frac{-6 + 0}{2})$$

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

Now, slope of AB = $$\frac{y_2 - y_1}{x_2 - x_1} = \frac{(0 + 6)}{(4 - 2)}$$

= $$\frac{6}{2} = 3$$

Let slope of line $$l = m$$

Product of slopes of two perpendicular lines = -1

=> $$m \times 3 = -1$$

=> $$m = \frac{-1}{3}$$

Equation of a line passing through point $$(x_1,y_1)$$ and having slope $$m$$ is $$(y - y_1) = m(x - x_1)$$

$$\therefore$$ Equation of line $$l$$

=> $$(y + 3) = \frac{-1}{3}(x - 3)$$

=> $$3y + 9 = -x + 3$$

=> $$x + 3y = 3 - 9 = -6$$

=> Ans - (B)

It is given that AD : DB = 1 : 2

Let AD = 1 cm and DB = 2 cm

Let area of $$\triangle$$ ADE = $$x$$ sq cm

In $$\triangle$$ ADE and $$\triangle$$ ABC

$$\angle$$ DAE = $$\angle$$ BAC  (common)

$$\angle$$ ADE = $$\angle$$ ABC   (Alternate interior angles)

$$\angle$$ AED = $$\angle$$ ACB    (Alternate interior angles)

=> $$\triangle$$ ADE $$\sim$$ $$\triangle$$ ABC

=> Ratio of Area of $$\triangle$$ ADE : Area of $$\triangle$$ ABC = Ratio of square of corresponding sides = $$(AD)^2$$ : $$(AB)^2$$

= $$\frac{(1)^2}{(1 + 2)^2} = \frac{x}{(45)}$$

=> $$\frac{x}{45} = \frac{1}{9}$$

=> $$x=\frac{45}{9}=5$$

$$\therefore$$ Area of quadrilateral BDEC = ar($$\triangle$$ ABC) - ar($$\triangle$$ ADE)

= $$45-5=40$$ $$cm^2$$

=> Ans - (B)

Coordinates of centroid of triangle with vertices $$(x_1 , y_1)$$ , $$(x_2 , y_2)$$ and $$(x_3 , y_3)$$ is $$(\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3})$$

Let coordinates of vertex C = $$(x , y)$$

Vertex A(7,-1) and Vertex B(1,2) and Centroid = (2,2)

=> $$2 = \frac{7 + 1 + x}{3}$$

=> $$x + 8 = 2 \times 3 = 6$$

=> $$x = 6 - 8 = -2$$

Similarly, => $$2 = \frac{-1 + 2 + y}{3}$$

=> $$y + 1 = 2 \times 3 = 6$$

=> $$y = 6 - 1 = 5$$

$$\therefore$$ Coordinates of vertex C = (-2,5)

=> Ans - (A)