Maths Solve the Following Question.(4 Marks)

Let the equation of the line be $\frac{x}{a}+\frac{y}{b}=1 \ldots \ldots$ (i)

This line passes through A(3, 5).

$\therefore \frac{3}{a}+\frac{5}{b}=1$

......(ii)

Since the required line makes equal intercepts on the co-ordinates axes,

a = b …….(iii)

Substituting the value of b in (ii), we get

$\begin{aligned} & \frac{3}{a}+\frac{5}{a}=1 \\ & \therefore a=8 \\ & \therefore b=8 \ldots \ldots[\text { From (iii)] }\end{aligned}$

Substituting the values of a and b in equation (i), the equation of the required line is

$\begin{aligned} & \frac{x}{8}+\frac{y}{8}=1 \\ & \therefore x+y=8\end{aligned}$

Case II: Line passing through origin. Slope of line passing through origin and A(3, 5) is

$m=\frac{5-0}{3-0}=\frac{5}{3}$

∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx. ∴ The equation of the required line is

$\begin{aligned} & y=\frac{5}{3} x \\ & \therefore 5 x-3 y=0\end{aligned}$

Slope of side $B C=\frac{2-6}{-1-7}=\frac{-4}{-8}=\frac{1}{2}$

∴ Slope of AD = – 2 [∵ AD ⊥ BC]

∴ Equation of line AD is y – (-2) = (- 2) (x – 3)

∴ y + 2 = -2x + 6

∴ 2x + y -4 = 0 …(i)

Slope of side $A C=\frac{-2-2}{3-(-1)}=\frac{-4}{4}=-1$

∴ Slope of BE = 1 …[ ∵ BE ⊥ AC]

∴ Equation of line BE is

y – 6 = 1(x – 7)

∴ y – 6 = x – 1

∴ x = y + 1 …(ii)

Substituting x = y + 1 in (i), we get

2(y + 1) + y – 4 = 0

∴ 2y + 2 + y – 4 = 0

∴ 3y – 2 = 0

$\therefore y=\frac{2}{3}$ in $($ ii $)$, wegetSubstitutingy $=[$ latex $] \frac{2}{3}$ in (ii), we get

$x=\frac{2}{3}+1=\frac{5}{3}$

$\therefore$ Co-ordinates of orthocentre, $O=\left(\frac{5}{3}, \frac{2}{3}\right)$

point A(- 2,3) to the line

3x-y- 1 = 0 …(i)

Slope of the line $3 x-y-1=0$ is $\frac{-3}{-1}=3$.

∴ Equation of AM is

$y-3=\frac{-1}{3}(x+2)$

∴ 3(y – 3) = – 1(x + 2) ∴ 3y – 9 = -x – 2

∴ x + 3y – 7 = 0 …………(ii)

The foot of perpendicular i.e., point M, is the point of intersection of equations (i) and (ii).

By (i) x 3 + (ii), we get 10x -10 = 0

∴ x = 1

Substituting x = 1 in (ii), we get

1 + 3y – 7 = 0

∴ 3y = 6

∴ y = 2

∴ The co-ordinates of the foot of the perpendicular Mare (1,2).

Slope of $O N=\frac{-4-0}{3-0}=\frac{-4}{3}$

Since line L ⊥ ON,

slope of the line $L$ is $\frac{3}{4}$ and it passes through point $N(3,-4)$.

Equation of the line in slope point form is $y-y_1=m\left(x-x_1\right)$

Equation of line L is

$y-(-4)=\frac{3}{4}(x-3)$

∴ 4(y + 4) = 3(x – 3)

∴ 4y + 16 = 3x – 9

∴ 3x – 4y – 25 = 0

Let the equation of the line be $\frac{x}{a}+\frac{y}{b}=1$

This line passes through A(3, 4).

$\therefore \frac{3}{a}+\frac{4}{b}=1$……………..(ii)

Since, the required line make equal intercepts on the co-ordinate axes.

∴ a = b …(iii)

Substituting the value of b in (ii), we get

$\begin{aligned} & \frac{3}{a}+\frac{4}{a}=1 \\ & \therefore \frac{7}{a}=1 \\ & \therefore a=7 \\ & \therefore b=7 \ldots[\text { From (iii)] }\end{aligned}$

Substituting the values of a and b in (i), equation of the required line is

$\frac{x}{7}+\frac{y}{7}=1=1$

∴ x + y = 7

Case II: Line passing through origin.

Slope of line passing through origin and $A(3,4)$ is $m==\frac{4-0}{3-0}=\frac{4}{3}$

∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.

∴ The equation of the required line is 4

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

∴ 4x – 3y = 0

Let the equation of the line be

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

Since, the sum of the intercepts of the line is zero.

∴ a + b = 0

∴ b = – a Substituting b = – a in (i), we get

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

x – y = a .. .(ii)

Since, the line passes through A(1, 3).

∴ 1 – 3 = a

∴ a = – 2 Substituting the value of a in (ii), equation of the required line is

∴ x – y = – 2,

∴ x – y + 2 = 0

Case II: Line passing through origin.

Slope of line passing through origin and

$A(1,3)$ is $m=\frac{3-0}{1-0}=3$

∴ Equation of the line having slope m and passing through origin (0, 0) is / = mx.

∴ The equation of the required line is y = 3x

∴ 3x – y = 0