Answer
$\text{a) Slope-Intercept Form: }
y=-\dfrac{5}{2}x+13
\\\\\text{b) Standard Form: }
5x+2y=26$
Work Step by Step
$\bf{\text{Solution Outline:}}$
Find the slope of the given equation, $
2x-5y=7
.$ Then use the negative reciprocal of this slope and the given point , $(
4,3
)$ to find the equation of the needed line. Give the equation in the Slope-Intercept Form and in the Standard Form.
$\bf{\text{Solution Details:}}$
In the form $y=mx+b$ or the Slope-Intercept Form (where $m$ is the slope), the given equation is equivalent to
\begin{array}{l}\require{cancel}
2x-5y=7
\\\\
-5y=-2x+7
\\\\
\dfrac{-5y}{-5}=\dfrac{-2x}{{-5}}+\dfrac{7}{-5}
\\\\
y=\dfrac{2}{5}x-\dfrac{7}{5}
.\end{array}
Hence, the slope of the given line is $
m=\dfrac{2}{5}
.$
Since perpendicular lines have negative reciprocal slopes, then the needed line has the following characteristics:
\begin{array}{l}\require{cancel}
m=-\dfrac{5}{2}
\\\text{Through: }
(4,3)
.\end{array}
Using $y-y_1=m(x-x_1)$ or the Point-Slope Form of linear equations, the equation of the line with the given conditions,
\begin{array}{l}\require{cancel}
y_1=3
,\\x_1=4
,\\m=-\dfrac{5}{2}
,\end{array}
is
\begin{array}{l}\require{cancel}
y-y_1=m(x-x_1)
\\\\
y-3=-\dfrac{5}{2}(x-4)
.\end{array}
Using the properties of equality, in the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y-3=-\dfrac{5}{2}(x)-\dfrac{5}{2}(-4)
\\\\
y-3=-\dfrac{5}{2}x+10
\\\\
y=-\dfrac{5}{2}x+10+3
\\\\
y=-\dfrac{5}{2}x+13
.\end{array}
Using the properties of equality, in the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=-\dfrac{5}{2}x+13
\\\\
2(y)=2\left( -\dfrac{5}{2}x+13 \right)
\\\\
2y=-5x+26
\\\\
5x+2y=26
.\end{array}
The equation of the line is
\begin{array}{l}\require{cancel}
\text{a) Slope-Intercept Form: }
y=-\dfrac{5}{2}x+13
\\\\\text{b) Standard Form: }
5x+2y=26
.\end{array}