Question

The graph of $y=5x+3$ is plotted on $xy$ plane Further it is shifted down by $4$ units. Calculate $x$-intercept of the new line.

• A. $-1$
• B. $\frac{1}{5}$
• C. $1$
• D. $\frac{5}{4}$

Answer 1

The correct option is B $\frac{1}{5}$

Given The line $y=5x+3$

put $x=0,y=3(0,3)$

put $x=1y=8(1,8)$

put $x=-1,y=-2(-1,-2)$

By plotting there points in the xy - plane we get a straight line

The intercept is shifter 4 units down.So new points are (0,-1) and (1,4)

mid point $=(\frac{1+0}{2},\frac{4-1}{2})=(\frac{1}{2},\frac{3}{2})$

Slope of the line $\frac{y_2-y_1}{x_2-x_1}=\frac{4+1}{1-0}=\frac{5}{1}$

$\therefore$ Slope of the perpendicular bisector or intercept $=\frac{-1}{5}$

$\therefore$ Equation of new line, $y-\frac{3}{2}=\frac{-1}{5}\left(x\frac{-1}{2}\right)$

$5(2y-3)=-2x+1$

$⟹2x+10-16=0⟹x+5y-8=0$