What is the equation of the line through the point (3, 4) and perpendicular to 7x + 5y – 10 =0?
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If a line is said to be perpendicular to a given line, its slope should be equal to the negative reciprocal of the given line.
The given line is 7x + 5y - 10 = 0. To get the slope of this given equation, we have to transform this equation in the slope-intercept form y = mx +b. The equation becomes,
5y = -7x + 10
y = (-7/5)x + 2
so the slope is -7/5.
So the slope of the new line perpendicular to the given line is 5/7 since it should be negative reciprocal, and it should pass the point (3,4)
Using point-slope form y - y1 = m(x - x1)
y - 4 = (5/7)(x - 3)
(The entire section contains 3 answers and 305 words.)
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Slope of a line
ax + by + c = 0 is given by:
Slope of line = -a/b
And slope of a line perpendicular to it is given by
Slope of perpendicular line = b/a
Thus the equation of the perpendicular line is:
bx - ay + c1 = 0
Substituting the given values of a and b in above line of equation the equation becomes:
5x - 7y + c1 = 0 ... (1)
o find the values of c1 for the perpendicular line passing through the point (3, 4) we substitute these coordinates in the above equation:
5*3 - 7*4 + c1 = 0
15 - 28 = - c1
c1 = 13
Substituting this value of c1 in equation (1) we get:
5x - 7y + 13 = 0
To find the the equation of the line through the point (3, 4) and perpendicular to 7x + 5y – 10 =0.
We know that the line perpendicular to ax+by+c = 0 is given by:
bx-ay +k = 0, where k is any constant to be determined by the given condition.
Therefore the line perpendicular to 7x+5y-10 = 0 is given by:
5x-7y+ k = 0....(1). Since this line also passes through the point (3,4), the coordinates of (3,4) should satisfy 5x-7y+k = 0.
So 5*3-7*4+k = 0.
15-28 +k = 0.
k = 28 -15 = 13.
Therefore substituting k= 1, in the equation at (1), we get:
5x-7y +13 = 0.
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