What is the equation of a line that passes through the point (2, 7) and is perpendicular to the line whose equation is y=x4+5 ?Enter your answer in the box.

Answer:[tex]y=-4x+15[/tex]Step-by-step explanation:The correct question isWhat is the equation of a line that passes through the point (2, 7) and is perpendicular to the line whose equation is y=(x/4)+5 ?step 1Find out the slope of the given linewe have[tex]y=\frac{1}{4}x+5[/tex]sothe slope is [tex]m=\frac{1}{4}[/tex]step 2Find out the slope of the line perpendicular to the given lineRemember thatIf two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)[tex]m_1*m_2=-1[/tex]we have[tex]m_1=\frac{1}{4}[/tex]substitute[tex]\frac{1}{4}*m_2=-1[/tex][tex]m_2=-4[/tex]step 3Find the equation of the lineThe equation of the line in point slope form is[tex]y-y_1=m(x-x_1)[/tex]we have[tex](x_1,y_1)=(2,7)[/tex][tex]m=-4[/tex]substitute[tex]y-7=-4(x-2)[/tex]Convert to slope intercept form[tex]y=mx+b[/tex][tex]y-7=-4x+8[/tex][tex]y=-4x+8+7[/tex][tex]y=-4x+15[/tex]