How do you find the slope and y intercept of the line that is perpendicular to #y=-x-1# and passes through the point (5, 7)?

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Answer 1 · 2016-11-26T11:29:49

The equation of the line is #y=x+2#
The slope is #=1#
The intercepts are #(0,2)# and #(-2,0)#

The equation of a line is #y=mx+c#

where #m# is the slope

The slope of the line #y=-x-1# is #m_1=-1#

The slope of the perpedicular line is #m_2#

#m_1*m_2=-1#

So #m_2=1#

The equation of a line, that passes through #(x_0,y_0)# with a slope of #m_2# is

#y-y_0=m_2(x-x_0)#

Here, #(x_0,y_0)=(5,7)#

So, the equation is

#y-7=1(x-5)#

#y=x+2#

graph{(y+x+1)(y-x-2)=0 [-7.9, 7.9, -3.95, 3.95]}