Question 143·Medium·Algebra

If one line has slope $m$, a line perpendicular to it has slope $-\tfrac{1}{m}$. Apply this to the slope you just found.

Write an equation for the new line using its slope and the point $(2,3)$ (for example, with point-slope form). Then, to get the $x$-intercept, what value should $y$ have, and how can you solve for $x$?

Type y = (4x - 7)/2 into Desmos to graph the line $4x - 2y = 7$ and visually confirm its positive slope.

In a new line, type y - 3 = -1/2 (x - 2) (or y = -1/2x + 4) to graph the line that is perpendicular to the first line and passes through $(2,3)$.

Look at where this perpendicular line crosses the $x$-axis (where $y = 0$). Click that intersection point and read off the $x$-coordinate and $y$-coordinate; this point is the $x$-intercept, which should match one of the answer choices.

Rewrite $4x - 2y = 7$ in slope-intercept form $y = mx + b$.

So the slope of this line is $2$.

For perpendicular lines, the slopes are negative reciprocals.

• The given line has slope $2 = \tfrac{2}{1}$.
• The negative reciprocal is $-\tfrac{1}{2}$.

So the slope of line $\ell$ is $-\tfrac{1}{2}$.

Line $\ell$ passes through $(2,3)$ and has slope $-\tfrac{1}{2}$, so use point-slope form:

Now solve for $y$ to get slope-intercept form:

This is the equation of line $\ell$.

The $x$-intercept is where the line crosses the $x$-axis, so $y = 0$.

Set $y = 0$ in $y = -\tfrac{1}{2}x + 4$:

So the $x$-intercept of line $\ell$ is $(8, 0)$.