Question

??? Can someone help me to Write the equation of the line which passes through (2, –3) and is perpendicular to y = 4x + 7 in standard form

Answer 1

do you know the relationship between perpendicular slopes?

Answer 2

no

Answer 3

:(( oh yes but what I have problems is with the formula and when I use it or not

Answer 4

ok... what is the slope we need...

Answer 5

the relationship between perpendicular slopes \(\large m_1 \) and \(\large m_2 \) is \(\large m_1\cdot m_2=-1 \)

Answer 6

so the slope is -1?

Answer 7

nope... what is the slope of the given line?

Answer 8

4 and reciprocal -1/4

Answer 9

so the slope we want is \(\large m=-\frac{1}{4} \)

Answer 10

now we have a point and a slope you can put it into point-slope form: \(\large y-y_0=m(x-x_0) \)

Answer 11

I used the formula I told you before and here is what I have so far\[y-(-3)=-1/4=(x-2) y+3=-1/4+1/2

Answer 12

that is were I got stuck

Answer 13

\(\large y-(-3)=-\frac{1}{4}(x-2) \)
\(\large y+3=-\frac{1}{4}x-(-\frac{1}{4})(2) \)

how about from here....

Answer 14

yes , after you distribute everything

Answer 15

notice on the right side you have two "negative" signs... they can change to "+":
\(\large y-(-3)=-\frac{1}{4}(x-2) \)
\(\large y+3=-\frac{1}{4}x-(-\frac{1}{4})(2) \)
\(\large y+3=-\frac{1}{4}x+(\frac{1}{4})(2) \)

Answer 16

\(\large y+3=-\frac{1}{4}x+\frac{1}{2} \)

Answer 17

yes that is what I have