Maths Solve the Following Question.(2 Marks)

Given, m = 3, c = 4

The equation of the line is y = 3x + 4

3x – y = -4

$\begin{aligned} & \frac{3 x}{(-4)}-\frac{y}{(-4)}=1 \\ & \frac{x}{\left(\frac{-4}{3}\right)}+\frac{y}{4}=1\end{aligned}$

This equation is of the form $\frac{x}{a}+\frac{g}{b}=1$, where

x-intercept = a

$\mathrm{x}$-intercept $=\frac{-4}{3}$

Alternate Method:

Let θ be the inclination of the line. Then tan θ = 3 …..[∵ slope = 3 (given)]

$\begin{aligned} & \frac{\mathrm{OB}}{\mathrm{OA}}=3 \\ & \frac{4}{\mathrm{OA}}=3 \\ & \mathrm{OA}=\frac{4}{3}\end{aligned}$

$x$-intercept $=-\frac{4}{3}$ as point $A$ is to the left side of $Y$-axis.

Since the required line is perpendicular to ST,

slope of required line $=\frac{-1}{3}$ and line passes through $A(-2,3)$

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

The equation of the required line is

$\begin{aligned} & y-3=\frac{-1}{3}(x+2) \\ & \Rightarrow 3(y-3)=-(x+2) \\ & \Rightarrow 3 y-9=-x-2 \\ & \Rightarrow x+3 y=7\end{aligned}$

∴ Distance between the parallel lines

$\begin{aligned} & =\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right| \\ & =\left|\frac{3-15}{\sqrt{3^2+4^2}}\right|=\left|\frac{-12}{\sqrt{9+16}}\right|\end{aligned}$

$=\frac{12}{5}$ units

$\begin{aligned} & \mathbf{p}=\left|\frac{\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right| \\ & \mathbf{p}=\left|\frac{78}{\sqrt{12^2+5^2}}\right|=\left|\frac{78}{\sqrt{144+25}}\right|\end{aligned}$

$\begin{aligned} & =\frac{78}{13} \text { } \\ & =6 \text { units }\end{aligned}$

$\begin{aligned} & \frac{3 x}{2}+\frac{2 y}{2}=1 \\ & \frac{x}{\frac{2}{3}}+\frac{y}{1}=1\end{aligned}$

This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$, with $a=\frac{2}{3}, b=1$.

The line $3 x+2 y=2$ intersects the $X$-axis at $A\left(\frac{2}{3}, 0\right)$ and $Y$-axis at $B(0,1)$.

Required line is passing through the midpoint of AB

Midpoint of $A B=\left(\frac{\frac{2}{3}+0}{2}, \frac{0+1}{2}\right)=\left(\frac{1}{3}, \frac{1}{2}\right)$

$\therefore$ Required line passes through $(0,0)$ and $\left(\frac{1}{3}, \frac{1}{2}\right)$.

Equation of the line in two point form is

$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

∴ The equation of the required line is

$\begin{aligned} & \frac{y-0}{\frac{1}{2}-0}=\frac{x-0}{\frac{1}{3}-0} \\ & 2 y=3 x \\ & \therefore 3 x-2 y=0\end{aligned}$

$\therefore$ It's equation is given by the formula
$
\begin{aligned}
& \left(y-y_1\right)=m\left(x-x_1\right) \\
& \therefore \quad(y-7)=-\frac{1}{3}(x-2) \\
& \therefore \quad 3 y-21=x+2 \\
& \therefore \quad x+3 y=23 .
\end{aligned}
$

The $x$ - intercept is $-\frac{\mathrm{c}}{\mathrm{a}}=-\frac{-15}{1}=15$
The $y-$ intercept is $-\frac{c}{b}=-\frac{-15}{3}=5$

$9 x+6 y-1=0$ i,e, $3 x+2 y-\frac{7}{3}=0$

Here, $a=3, b=2, c_1=6$ and $c_2=\frac{-7}{3}$

∴ Distance between the parallel lines

$=\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|=\left|\frac{6-\left(-\frac{7}{3}\right)}{\sqrt{3^2+2^2}}\right|$

$=\left|\frac{6+\frac{7}{3}}{\sqrt{9+4}}\right|=\frac{25}{3 \sqrt{13}}$ units

Here, a = 4, b = – 3, c1 = 5 and c2 = 7

∴ Distance between the parallel lines

$\begin{aligned} & =\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right|=\left|\frac{5-7}{\sqrt{4^2+(-3)^2}}\right| \\ & =\left|\frac{-2}{\sqrt{16+9}}\right|=\left|\frac{-2}{5}\right|=\frac{2}{5} \text { units }\end{aligned}$

$\therefore \quad \mathrm{p}=\left|\frac{\mathrm{ar}+\mathrm{b} y_1+\mathrm{c}}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right|=\left|\frac{12(-2)-5(3)-13}{\sqrt{12^2+(-5)^2}}\right|$

$=\left|\frac{-24-15-13}{\sqrt{144+25}}\right|=\left|\frac{-52}{13}\right|=4$ units

7x + 24y – 50 = 0

Here, a = 7, b = 24, c = -50

$\begin{aligned} & \therefore \quad p=\left|\frac{c}{\sqrt{a^2+b^2}}\right| \\ & \therefore \quad p=\left|\frac{-50}{\sqrt{7^2+24^2}}\right|=\left|\frac{-50}{\sqrt{49+576}}\right|=\left|\frac{-50}{25}\right|=2 \text { units }\end{aligned}$

$3 x-y+23=0$ is $\frac{-1}{3}$

Since the x-intercept of the required line is 3, it passes through (3, 0).

∴ The equation of the required line is ‘

$y-0=\frac{-1}{3}(x-3)$

∴ 3y = x + 3

∴ x + 3y = 3

$\therefore m_1=\frac{- \text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}$

Let $m_2$ be the slope of the line $2 x-4 y+15=0$.

$\therefore m_2=\frac{- \text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{-4}=\frac{1}{2}$

Since m1 = m2 the given lines are parallel to each other.

Comparing this equation with ax + by + c = 0,

we get

a = 1, b = 2, c = 0

$\begin{array}{r}\therefore \quad \text { Slope of the line }=\frac{-a}{b}=\frac{-1}{2} \\ x \text {-intercept }=\frac{-c}{a}=\frac{-0}{1}=0 \\ y \text {-intercept }=\frac{-c}{b}=\frac{-0}{2}=0\end{array}$

Comparing this equation with ax + by + c = 0,

we get

a = 3, b = – 1, c = – 9

$\begin{aligned} & \therefore \quad \text { Slope of the line }=\frac{-a}{b}=\frac{-3}{-1}=3 \\ & x \text {-intercept }=\frac{-c}{a}=\frac{-(-9)}{3}=3 \\ & y \text {-intercept }=\frac{-c}{b}=\frac{-(-9)}{-1}=-9\end{aligned}$

Comparing this equation with ax + by + c = 0,

we get

a = 2, b = 3, c = -6

$\therefore \quad$ Slope of the line $=\frac{-\mathrm{a}}{\mathrm{b}}=\frac{-2}{3}$

$x$-intercept $=\frac{-c}{a}=\frac{-(-6)}{2}=3$

$y$-intercept $=\frac{-\mathrm{c}}{\mathrm{b}}=\frac{-(-6)}{3}=2$

$\begin{aligned} \therefore \quad \mathrm{D} & =\left(\frac{3+2}{2}, \frac{4+0}{2}\right)=\left(\frac{5}{2}, 2\right) \text { and } \\ \mathrm{E} & =\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\end{aligned}$

$\therefore \quad$ The equation of the line $\mathrm{DE}$ is

$\begin{aligned} \frac{y-2}{3-2} & =\frac{x-\frac{5}{2}}{\frac{1}{2}-\frac{5}{2}} \\ \therefore \quad \frac{y-2}{1} & =\frac{2 x-5}{-4}\end{aligned}$

∴ -4(y-2) = 2x-5

∴ 2x + 4y – 13 = 0

$\therefore D=\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)$

The median AD passes through the points

$A(3,4)$ and $D\left(\frac{1}{2}, 3\right)$

∴ The equation of the median AD is

$\begin{aligned} & \frac{y-4}{3-4}=\frac{x-3}{\frac{1}{2}-3} \\ & \frac{y-4}{-1}=\frac{x-3}{-\frac{5}{2}}\end{aligned}$

$\frac{5}{2}(y-4)=x-3$

∴ 5y – 20 = 2x – 6

∴ 2x – 5y + 14 = 0

Equation of the line in two point form is

$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

∴ The equation of the side BC is

$\begin{aligned} & \frac{y-0}{6-0}=\frac{x-2}{-1-2} \\ & \frac{y}{6}=\frac{x-2}{-3}\end{aligned}$

∴ – 3y = 6x – 12

∴ 6x + 3y – 12 = 0

∴ 2x + y – 4 = 0

∴ Slope of the line (m) = tan 0 = tan 135°

= tan (90° + 45°)

= – cot 45° = – 1 x-intercept of the required line is 7.

∴ The line passes through (7, 0).

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

$\therefore$ The equation of the required line is $y-0=-1(x-7)$

∴ y = -x + 7

∴ x + y – 7 = 0

Equation of the line in two point form is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

∴ The equation of the required line is

$\begin{aligned} & \frac{y-1}{2-1}=\frac{x-2}{3-2} \\ & \therefore \frac{y-1}{1}=\frac{x-2}{1} \\ & \therefore y-1=x-2 \\ & \therefore y=x-1\end{aligned}$

Comparing this equation with y = mx + c,

we get m = 1 and c = – 1

Alternate Method:

Points A(2, 1) and B(3, 2) lie on the line y = mx + c.

∴ They must satisfy the equation.

∴ 2m + c = 1 …(i) and 3m + c = 2 …(ii) equation (ii) – equation (i) gives m = 1

Substituting m = 1 in (i), we get 2(1) + c = 1

∴ c = 1 – 2 = – 1

Given equation of the line is 3x +y = 6.

$\therefore \frac{x}{2}+\frac{y}{6}=1$

This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$,

where a = 2, b = 6 ∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2, 0) and B(0, 6) respectively. Required line is passing through the midpoint of AB.

$\therefore$ Midpoint of $A B=\left(\frac{2+0}{2}, \frac{0+6}{2}\right)=(1,3)$

∴ Required line passes through (0, 0) and (1,3).

Equation of the line in two point form is

$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

∴ The equation of the required line is

$\begin{aligned} & \frac{y-0}{3-0}=\frac{x-0}{1-0} \\ & \frac{y}{3}=\frac{x}{1} \\ & \therefore y=3 x \\ & \therefore 3 x-y=0\end{aligned}$

Alternate Method:

Given equation of the line is 3x + y = 6 …

(i) Substitute y = 0 in (i) to get a point on X-axis.

∴ 3x + 0 = 6

∴ x = 2 Substitute x = 0 in

(i) to get a point on Y-axis.

∴ 3(0) + 7 = 6 ∴ y = 6

∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2,0) and B(0,6) respectively.

Let M be the midpoint of AB.

$M=\left(\frac{2+0}{2}, \frac{0+6}{2}\right)=(1,3)$

Slope of $O M(m)=\frac{3-0}{1-0}=3$

Equation of OM is of the formy = mx.

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

∴ 3x – y = 0

$=-\sqrt{3}$

and the line passes through A(4, 3).

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

∴ The equation of the required line is

$\begin{aligned} & y-3=-\sqrt{3}(x-4) \\ & \therefore y-3=-\sqrt{3} x+4 \sqrt{3} \\ & \therefore \sqrt{3} x+y-3-4 \sqrt{3}=0\end{aligned}$

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

$\therefore$ The equation of the required line is $y-5=\frac{2}{3}(x-3)$

∴ 3 (y – 5) = 2 (x – 3)

∴ 3y – 15 = 2x – 6

∴ 2x – 3y + 9 = 0

Equation of the line in slope point form is

$y-y_1=m\left(x-x_1\right)$

∴ The equation of the required line is

$\begin{aligned} & {[y-(-2)]=\frac{1}{2}(x-3)} \\ & \therefore 2(y+2)=x-3 \\ & \therefore 2 y+4=x-3 \\ & \therefore x-2 y-7=0\end{aligned}$

Slope of $A B=\frac{7-4}{1-2}=-31-2$

Since the required line is parallel to line AB, slope of required line (m) = slope of AB

∴ m = – 3 and the required line passes through the origin.

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

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

Slope of the line (m) = tan θ = tan 60°

$=\sqrt{3}$

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

$\therefore$ The equation of the required line is $y=\sqrt{3} \mathrm{x}$