Find the intersection of the line L through (1, 2) and (5, 8) with the line M through (2,2) and (4, 0).
The slope of L is (8 — 2)/(5 — 1) = |. Its slope-intercept equation has the form y = \x + b. Since it
passes through (1,2), 2=|(l) + fe, and, therefore, b=\. So, L has equation y=\x+\. Similarly,passes through (1,2), 2=|(l) + fe, and, therefore, b=\. So, L has equation y=\x+\. Similarly,
we find that the equation of M is y = -x + 4. So, we must solve the equations y = -x + 4 and y = \x + \
simultaneously. Then, -x + 4=\x+\, -2x + 8 = 3x + 1, 7 = 5*, x=\. When x=l, y = -x +
4=-s+4=T- Hence, the point of intersection is (|, " )•
Find the distance from the point (1, 2) to the line 3x - 4y = 10.
Remember that the distance from a point (*,, y : ) to a line Ax + By+C = 0 is \Ax l + By l + C\l
^A
2 + B
2 . In our case, A = 3, B = -4, C=10, and VA
2 + B
2 = V25 = 5. So, the distance is
|3(l)-4(2)-10|/5=^=3.
Find equations of the lines of slope — | that form with the coordinate axes a triangle of area 24 square units.
The slope-intercept equations have the form y = - \x + b. When y = 0, x = 56. So, the x-intercept a
is 56. Hence, the area of the triangle is \ab = \(%b)b = \b
2 = 24. So, fo
2 = 36, b = ±6, and the desired
equations are y=-\x± 6; that is, 3* + 4y = 24 and 3x + 4y = -24.
A point (x, y) moves so that its distance from the line x = 5 is twice as great as its distance from the line
y = 8. Find an equation of the path of the point.
The distance of (x, y) from x = 5 is |jc-5|, and its distance from y = 8 is |y-8|. Hence,
|x-5|=2|y-8|. So, x - 5 = ±2(y - 8). ' There are two cases: x- 5 = 2(^-8) and x-5=-2(y-8),
yielding the lines x-2y = -ll and * + 2>> = 21. A single equation for the path of the point would
be (x-2y + ll)(x + 2y-2l) = 0.
Find the equations of the lines through (4, —2) and at a perpendicular distance of 2 units from the origin.
A point-slope equation of a line through (4, —2) with slope m is y + 2 = m(x — 4) or mx — y — (4m +
2) = 0. The distance of (0, 0) from this line is |4m + 2| A/m
2 + 1. Hence, |4m + 2| /V'm
2 + 1 = 2. So,
(4/n + 2)
2 = 4(w
2 + l), or (2m +1)
2 = m
2 +1. Simplifying, w(3m + 4) = 0, and, therefore, m = 0 or
OT = - 5. The required equations are y = -2 and 4x + 3y - 10 = 0.
In Problems 3.49-3.51, find a point-slope equation of the line through the given points.
(2,5) and (-1,4).
m = (5-4)/[2-(-l)]=|. So, an equation is (y - 5)/(x -2) = £ or y-5=$(*-2).
(1,4) and the origin.
m = (4 — 0)/(1 — 0) = 4. So, an equation is y/x = 4 or y = 4x.
(7,-1) and (-1,7).
m = (-l-7)/[7-(-l)] = -8/8=-l. So, an equation is (y + l)/(x -7) = -1 or y + l = -(x-l).
In Problems 3.52-3.60, find the slope-intercept equation of the line satisfying the given conditions.
Through the points (-2,3) and (4,8).
w = (3-8)/(-2-4)=-5/-6= §. The equation has the form y=\x+b. Hence, 8=i(4) + Z>,
b = ". Thus, the equation is y = \x + ".
Having slope 2 and y-intercept — 1.
y = 2x-l.
Through (1,4) and parallel to the x-axis.
Since the line is parallel to the jc-axis, the line is horizontal. Since it passes through (1, 4), the equation is
y = 4.
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The slope of L is (8 — 2)/(5 — 1) = |. Its slope-intercept equation has the form y = \x + b. Since it
passes through (1,2), 2=|(l) + fe, and, therefore, b=\. So, L has equation y=\x+\. Similarly,passes through (1,2), 2=|(l) + fe, and, therefore, b=\. So, L has equation y=\x+\. Similarly,
we find that the equation of M is y = -x + 4. So, we must solve the equations y = -x + 4 and y = \x + \
simultaneously. Then, -x + 4=\x+\, -2x + 8 = 3x + 1, 7 = 5*, x=\. When x=l, y = -x +
4=-s+4=T- Hence, the point of intersection is (|, " )•
Find the distance from the point (1, 2) to the line 3x - 4y = 10.
Remember that the distance from a point (*,, y : ) to a line Ax + By+C = 0 is \Ax l + By l + C\l
^A
2 + B
2 . In our case, A = 3, B = -4, C=10, and VA
2 + B
2 = V25 = 5. So, the distance is
|3(l)-4(2)-10|/5=^=3.
Find equations of the lines of slope — | that form with the coordinate axes a triangle of area 24 square units.
The slope-intercept equations have the form y = - \x + b. When y = 0, x = 56. So, the x-intercept a
is 56. Hence, the area of the triangle is \ab = \(%b)b = \b
2 = 24. So, fo
2 = 36, b = ±6, and the desired
equations are y=-\x± 6; that is, 3* + 4y = 24 and 3x + 4y = -24.
A point (x, y) moves so that its distance from the line x = 5 is twice as great as its distance from the line
y = 8. Find an equation of the path of the point.
The distance of (x, y) from x = 5 is |jc-5|, and its distance from y = 8 is |y-8|. Hence,
|x-5|=2|y-8|. So, x - 5 = ±2(y - 8). ' There are two cases: x- 5 = 2(^-8) and x-5=-2(y-8),
yielding the lines x-2y = -ll and * + 2>> = 21. A single equation for the path of the point would
be (x-2y + ll)(x + 2y-2l) = 0.
Find the equations of the lines through (4, —2) and at a perpendicular distance of 2 units from the origin.
A point-slope equation of a line through (4, —2) with slope m is y + 2 = m(x — 4) or mx — y — (4m +
2) = 0. The distance of (0, 0) from this line is |4m + 2| A/m
2 + 1. Hence, |4m + 2| /V'm
2 + 1 = 2. So,
(4/n + 2)
2 = 4(w
2 + l), or (2m +1)
2 = m
2 +1. Simplifying, w(3m + 4) = 0, and, therefore, m = 0 or
OT = - 5. The required equations are y = -2 and 4x + 3y - 10 = 0.
In Problems 3.49-3.51, find a point-slope equation of the line through the given points.
(2,5) and (-1,4).
m = (5-4)/[2-(-l)]=|. So, an equation is (y - 5)/(x -2) = £ or y-5=$(*-2).
(1,4) and the origin.
m = (4 — 0)/(1 — 0) = 4. So, an equation is y/x = 4 or y = 4x.
(7,-1) and (-1,7).
m = (-l-7)/[7-(-l)] = -8/8=-l. So, an equation is (y + l)/(x -7) = -1 or y + l = -(x-l).
In Problems 3.52-3.60, find the slope-intercept equation of the line satisfying the given conditions.
Through the points (-2,3) and (4,8).
w = (3-8)/(-2-4)=-5/-6= §. The equation has the form y=\x+b. Hence, 8=i(4) + Z>,
b = ". Thus, the equation is y = \x + ".
Having slope 2 and y-intercept — 1.
y = 2x-l.
Through (1,4) and parallel to the x-axis.
Since the line is parallel to the jc-axis, the line is horizontal. Since it passes through (1, 4), the equation is
y = 4.
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