How do you write an equation in standard form for a line through (1,-5); slope $-\dfrac{3}{2}$ ?

Hint: The standard form of a straight line is $y=mx+c$ where m is the slope of the line and c is the y intercept of the line. Slope of the line is given in the question and the intercept is not given , but we can find it by the given one point in the question. The given one point will satisfy the equation of line.

Complete step by step answer:
The slope of line is $-\dfrac{3}{2}$
We know that we can write any line as $y=mx+c$ and this is the standard form
So the equation of line is $y=-\dfrac{3}{2}x+c$
The given point which passes through the line is (1,-5)
Now we can find the value of c by using the given point , the point (1,-5) will satisfy the equation of line
So $-5=-\dfrac{3}{2}\times 1+c$
So the value of c is $\dfrac{3}{2}-5=-\dfrac{7}{2}$
Now we can write the equation of line is $y=-\dfrac{3}{2}x-\dfrac{7}{2}$

Note:
Another way to solve this problem to take another point on the line is (x,y) , so the line joining (x,y) and the given point (1,-5) will be equal to $-\dfrac{3}{2}$ so we can write
$\dfrac{y-\left( -5 \right)}{x-1}=-\dfrac{3}{2}$
Further solving by cross multiplication we get
$\Rightarrow 2y+10=3-3x$
$\Rightarrow 2y=-3x-7$
$\Rightarrow y=-\dfrac{3}{2}x-\dfrac{7}{2}$
We can see that the above solution is exactly the same as the first solution. We can see that the slope is equal to $-\dfrac{3}{2}$ and (1,-5) will satisfy the equation.