Question

triangular park has two vertices as B (-4, 1) and C (2, 11). The third vertex A is a point diving the line joining the points (3, 1) and (6, 4) in the ratio 2 : 1. Based on the above information, answer the following questions: 16. The coordinates of third vertex A are (a) (5, 3) (b) (3, 5) (c) (-5, 3) (d) (5, -3) 17. The equation passing through B and C is (a) 5x – 3y – 23 = 0 (b) 5x – 3y + 23 = 0 (c) 3x + 5y – 23 = 0 (d) 5x + 3y – 23 = 0 18. The equation passing through A and parallel to BC is (a) 5x – 3y + 16 = 0 (b) 5x – 3y + 34 = 0 (c) 5x – 3y - 16 = 0 (d) 5x + 3y – 16 = 0 19. The equation passing through A and perpendicular to BC is (a) 3x + 5y – 30 = 0 (b) 3x + 5y + 30 = 0 (c) 3x - 5y + 30 = 0 (d) 3x - 5y = 0 20. The area of triangular field ABC is (a) 78 sq. units (b) 43 sq. units (c) 86 sq. units (d) 39 sq. units


FIRST ANSWER WILL BE BRAINLEIST ANSWER​

Answer 1

Answer:

please tell properly

Answer 2

Given:

Coordinates of point B = (-4,1)

Coordinates of point C = (2,11)

Ratio in which point A divides the line = 2:1

Coordinates of the points joining the line = (3,1) and (6,4)

To find:

Coordinates of point A.

Equation of line passing through B and C.

Equation of line passing through A and parallel to BC.

Equation of line passing through A and perpendicular to BC.

Area of the triangular field ABC.

Solution:

By section formula, if a point having coordinates $(x,y)$ divides the line joining the points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ in a ratio $m:n$, then,

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

Given that point $A(x,y)$ cuts the line joining the points (3,1) and (6,4) in the ratio 2:1. Let these points be P(3,1) and Q(6,4).

$x_{1}=3,y_{1}=1,x_{2}=6,y_{2}=4$

$m=2,n=1$

$(x,y)=(\frac{2*6+1*3}{2+1} ,\frac{2*4+1*1}{2+1} )$

$(x,y)=(\frac{12+3}{3}, \frac{8+1}{3})$

$(x,y)=(\frac{15}{3},\frac{9}{3})$

$(x,y)=(5,3)$

Hence, point A has coordinates (5,3). Option (a) is the correct answer.

The slope of a line passing through two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $m$ and it is given by:

$m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

We calculate the slope of the line passing through points B(-4,1) and C(2,11).

$x_{1}=-4,y_{1}=1,x_{2}=2,y_{2}=11$

$m=\frac{11-1}{2-(-4)}$

$m=\frac{10}{6}$

∴ $m=\frac{5}{3}$

Using this slope and either of the two points to calculate the y-intercept.

$y=mx+b$

where, $b$ is the y-intercept

Using C(2,11)

$11=\frac{5}{3}*2+b$

$11-\frac{10}{3}=b$

$\frac{33-10}{3}=b$

∴ $b=\frac{23}{3}$

The equation in slope-intercept form is given by:

$y=mx+b$

$y=\frac{5}{3}x+\frac{23}{3}$

Multiplying by 3 on both sides,

$3y=5x+23$

$5x-3y+23=0$

Hence, the equation of the line passing through points B (-4,1) and C(2,11) is $5x-3y+23=0$. Option (b) is the correct answer.

The slope of the line joining the points B and C have been calculated as $\frac{5}{3}$. To determine the equation of a line passing through a point $(x,y)$ and parallel to another line having a slope $m$ is given by:

$y-y_{1}=m(x-x_{1})$

Here, we have point A(5,3) and we need to find the equation of a line passing through A and parallel to line BC. The equation of line BC has been determined previously.

$x_{1}=5,y_{1}=3,m=\frac{5}{3}$

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

$3(y-3)=5(x-5)$

$3y-9=5x-25$

$5x-3y-16=0$

Hence, the equation of a line passing through A and parallel to line BC is $5x-3y-16=0$. Option (c) is the correct answer.

When two lines are perpendicular, then the product of their slopes is -1.

$m_{1}m_{2}=-1$

Here, the slope of line BC is calculated as $m_{1}=\frac{5}{3}$

$\frac{5}{3}m_{2}=-1$

$m_{2}=\frac{-3}{5}$

Hence, the equation of a line passing through A and perpendicular to BC has a slope of $m_{2}=\frac{-3}{5}$

This equation passes through A(5,3). By slope-intercept form, determine the y-intercept.

$y=mx+b$

$y=3,x=5,m=\frac{-3}{5}$

$3=\frac{-3}{5}*5+b$

$3=-3+b$

∴ $b=3+3=6$

Thus, the required equation of the line is:

$y=\frac{-3}{5}x+6$

Multiplying by 5 on both sides,

$5y=-3x+30$

$3x+5y-30=0$

Hence, the equation of a line passing through A and perpendicular to BC is $3x+5y-30=0$. Option (a) is the correct answer.

The area of the triangular field whose coordinates are given as A(5,3), B(-4,1) and C(2,11) is determined using the formula,

$A=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

$x_{1}=5,y_{1}=3,x_{2}=-4,y_{2}=1, x_{3}=2,y_{3}=11$

Substituting in the above formula,

$A=\frac{1}{2}[5(1-11)-4(11-3)+2(3-1)]$

$A=\frac{1}{2}[5(-10)-4(8)+2(2)]$

$A=\frac{1}{2}(-50-32+4)$

$A=\frac{78}{2}=39$ sq.units

Hence, the area of triangular field ABC is 39 sq. units. Option (d) is the correct answer.

Coordinates of point A is (5,3). Option (a) is the correct answer.

Equation of line passing through B and C is $5x-3y+23=0$. Option (b) is the correct answer.

Equation of line passing through A and parallel to BC is $5x-3y-16=0$. Option (c) is the correct answer.

Equation of line passing through A and perpendicular to BC is $3x+5y-30=0$. Option (a) is the correct answer.

The area of the triangular field ABC is 39 sq. units. Option (d) is the correct answer.