Chapter Solution
Chapter-4
Question 1
A point P lies in the III quadrant. It is reflected across the y-axis to create P′ and then P′ is reflected across the x-axis to create P″. In which quadrants do P′ and P″ lie?
In the III quadrant, both coordinates are negative. Reflection in the y-axis changes the sign of the x-coordinate, so the point moves to the IV quadrant. Reflection of this point in the x-axis changes the sign of the y-coordinate, so it moves to the I quadrant.
Answer: P′ lies in Quadrant IV and P″ lies in Quadrant I.
Question 2
A point A(a, b) is reflected across the x-axis to become point B. Point B is exactly 8 units below point A. What was the y-coordinate of point A?
Reflection in the x-axis changes (a, b) to (a, −b). The vertical distance between the two points is |2b| = 8.
Since B is below A, A must lie above the x-axis, so b > 0. Therefore, b = 4.
Answer: The y-coordinate of A is 4.
Question 1
Point P(a, b) lies in IV Quadrant. Its perpendicular distance from the y-axis is 3 units greater than its perpendicular distance from the x-axis. If the product of its coordinates is −28, find a2 + b.
Since P lies in the IV quadrant, a > 0 and b < 0. The perpendicular distances from the y-axis and x-axis are a and −b respectively.
Thus, a = −b + 3. Also, ab = −28.
Substituting a = 3 − b gives b2 − 3b − 28 = 0.
Hence, (b − 7)(b + 4) = 0. Since b < 0, b = −4 and therefore a = 7.
Therefore, a2 + b = 49 + 4 = 45.
Question 2
Point Q(k − 4, 2k + 7) lies strictly in II Quadrant. The perpendicular distance of Q from the y-axis is exactly twice its perpendicular distance from the x-axis. Find k and k3 + 10.
In the II quadrant, k − 4 < 0 and 2k + 7 > 0. The distances are 4 − k and 2k + 7.
Hence, 4 − k = 2(2k + 7).
So, 4 − k = 4k + 14, giving −5k = 10 and k = −2.
Therefore, k3 + 10 = −8 + 10 = 2.
Question 3
Let P(3, −4) be a point on the Cartesian plane. Point Q(a, b) is the image of P when reflected through y-axis. Point R(c, d) is the image of Q when reflected in the x-axis. Find ac − bd.
Reflection of P(3, −4) in the y-axis gives Q(−3, −4), so a = −3 and b = −4. Reflection in the x-axis gives R(−3, 4), so c = −3 and d = 4.
Thus, ac − bd = (−3)(−3) − (−4)(4) = 9 + 16 = 25.
Question 1
A straight line with a slope of 1/2 passes through A(x, 2), B(3, 4) and C(7, y). Find A, C and x + y.
Using A and B, (4 − 2)/(3 − x) = 1/2, so 4 = 3 − x and x = −1.
Using B and C, (y − 4)/(7 − 3) = 1/2, so y − 4 = 2 and y = 6.
Answer: A = (−1, 2), C = (7, 6), and x + y = 5.
Question 2
Points P(2, 3), Q(5, 7), and R(13, k) are three consecutive vertices of a rectangle. Find k.
PQ is one side and QR is the next side, so they are perpendicular.
Slope(PQ) = 4/3 and slope(QR) = (k − 7)/8.
Therefore, (4/3)((k − 7)/8) = −1, giving k − 7 = −6.
Answer: k = 1.
Question 3
A line l passes through the origin. It intersects the line segment joining A(2, 8) and B(6, 2). What is the possible range of the slope m?
The slopes of OA and OB are 8/2 = 4 and 2/6 = 1/3. A ray from the origin that meets the segment AB must have a slope between these two limiting slopes.
Answer: 1/3 ≤ m ≤ 4.
Question 4
Line l1 passes through A(1, k) and B(k, 7). Line l2 passes through C(−2, 4) and D(1, 10). If l1 ∥ l2, find k and k2 + 5.
Slope of l2 = (10 − 4)/(1 + 2) = 2.
Slope of l1 = (7 − k)/(k − 1).
For parallel lines, these slopes are equal:
(7 − k)/(k − 1) = 2 ⇒ 7 − k = 2k − 2 ⇒ 3k = 9.
Hence k = 3 and k2 + 5 = 14.
Question 5
A line p has slope 2. A line q passes through M(3a, 4) and N(a, 5). The two lines are perpendicular. Find a, M and N.
Slope of q = (5 − 4)/(a − 3a) = −1/(2a).
For perpendicular lines, 2 × (−1/(2a)) = −1, so a = 1.
Answer: a = 1, M = (3, 4), N = (1, 5).
Question 1
The line kx + 3y − 12 = 0 forms a right-angled triangle with the axes. If the area is 12 square units, find all possible slopes.
x-intercept = 12/k and y-intercept = 4. The triangle area is 1/2 × |12/k| × 4 = 12, giving |k| = 2.
Since slope = −k/3, the possible slopes are −2/3 and 2/3.
Question 2
The line px + qy + r = 0 (p, q, r ≠ 0) forms an isosceles right-angled triangle with the coordinate axes in the first quadrant. Find the relation between p and q.
The x- and y-intercepts are −r/p and −r/q. For an isosceles right triangle these intercept lengths are equal, so −r/p = −r/q.
Since r ≠ 0, p = q.
Question 3
l1: 3x − 5y + 10 = 0 and l2: 5x + 3y + K = 0. Prove they are perpendicular. Also find K if the x-intercept of l1 equals the y-intercept of l2.
Slopes are m1 = 3/5 and m2 = −5/3. Their product is −1, so the lines are perpendicular.
The x-intercept of l1 is found from 3x + 10 = 0, giving x = −10/3.
The y-intercept of l2 is −K/3. Equating gives −K/3 = −10/3, hence K = 10.
Question 4
A line is written as kx − y + C = 0. It passes through (3, 10), and the sum of its x-intercept and y-intercept equals its slope. Find all possible equations.
From the point (3, 10), 3k − 10 + C = 0, so C = 10 − 3k.
The slope is k. The intercepts are −C/k and C. Hence their sum is C(1 − 1/k).
Setting this equal to k gives C(k − 1) = k2. Substituting C = 10 − 3k gives:
(10 − 3k)(k − 1) = k2 ⇒ 4k2 − 13k + 10 = 0.
Thus (4k − 5)(k − 2) = 0, giving k = 5/4 or k = 2.
For k = 2, C = 4, so the line is 2x − y + 4 = 0.
For k = 5/4, C = 25/4, so the line is 5x − 4y + 25 = 0.
Question 1
A line crosses the y-axis at (0, −5) and rises 3 units for every 2 units it moves right. Find its equation.
The slope is 3/2 and the y-intercept is −5.
Therefore, y = (3/2)x − 5.
Question 2
Find the equation of the line parallel to y = 5x − 12 and passing through (0, 9).
A parallel line has the same slope, 5. Since it passes through (0, 9), its intercept is 9.
Answer: y = 5x + 9.
Question 3
Find the equation of the line perpendicular to y = x/3 + 4 and passing through the origin.
The given slope is 1/3, so the perpendicular slope is −3.
Through the origin, the equation is y = −3x.
Question 4
A straight line passes through (0, −6) and (4, 10). Find its equation in slope-intercept form.
Slope = (10 + 6)/(4 − 0) = 16/4 = 4, and the y-intercept is −6.
Answer: y = 4x − 6.
Question 5
Convert 8x − y + 7 = 0 into slope-intercept form.
Rearranging, y = 8x + 7.
Answer: y = 8x + 7.
Question 1
A straight line passes through (2, 3) and forms a triangle with the positive axes of area 12 square units. Find its x-intercept a and y-intercept b.
Using intercept form, 2/a + 3/b = 1. Also, 1/2 ab = 12, so ab = 24.
Substituting b = 24/a gives 2/a + a/8 = 1, or a2 − 8a + 16 = 0.
Thus a = 4 and b = 6.
Answer: a = 4, b = 6.
Question 2
A straight line passes through (2, 2). The sum of its x-intercept a and y-intercept b is 9. Find ab.
Intercept form gives 2/a + 2/b = 1, so (a + b)/ab = 1/2.
Since a + b = 9, 9/ab = 1/2, hence ab = 18.
Question 3
A straight line passes through (3, 5). The sum of its x-intercept and y-intercept is zero. Find all possible equations.
For non-zero intercepts, let b = −a. Then x/a + y/(−a) = 1, or x − y = a. Substituting (3, 5) gives a = −2, hence y = x + 2.
There is also the line through the origin and (3, 5): its slope is 5/3, giving 5x − 3y = 0.
Answer: y = x + 2 and 5x − 3y = 0.
Question 4
A line forms a right-angled triangle with the positive axes. Its area is 24 square units and its hypotenuse is 10 units. Find all possible equations in intercept form.
Let the intercepts be a and b. Then ab/2 = 24, so ab = 48. Also, a2 + b2 = 100.
Thus (a + b)2 = 100 + 96 = 196, so a + b = 14.
The numbers a and b are therefore 6 and 8. Hence the possible equations are:
x/6 + y/8 = 1 and x/8 + y/6 = 1.
Question 5
A line passes through (3, 2). Its positive x- and y-intercepts have sum 12. Find all possible equations.
Let the intercepts be a and b. Then 3/a + 2/b = 1 and a + b = 12.
Putting b = 12 − a gives a2 − 13a + 36 = 0, so a = 9 or 4. Hence b = 3 or 8.
Therefore, the equations are x/9 + y/3 = 1 and x/4 + y/8 = 1, equivalently x + 3y = 9 and 2x + y = 8.
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