OP Malhotra Class 11 Maths Solutions Chapter 16 The Straight Line Chapter Test

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S Chand Class 11 ICSE Maths Solutions Chapter 16 The Straight Line Chapter Test

Question 1.
The coordinates of the points A, B, C are (0, 4),(2, 5) and (3, 3) respectively. Prove that AB = BC and the angle ABC is a right angle.
Solution:
The coordinates of points A, B and C are (0, 4), (2, 5) and (3, 3).
Here AB = \(\sqrt{(2-0)^2+(5-4)^2}\)
= \(\sqrt{4+1}\) = \(\sqrt{5}\)
BC = \(\sqrt{(3-2)^2+(3-5)^2}\) = \(\sqrt{1+4}\) = \(\sqrt{5}\)
∴ AB = BC and CA = \(\sqrt{(3-0)^2+(3-4)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)
Further AB² + BC² = 5 + 5 = 10 = CA²
Thus △ABC be right angled at point B.

Question 2.
Find the coordinates of the point which divides the line joining (5, – 2) and (9, 6) in the ratio 3 : 1.
Solution:
Let the point P divides the line segment AB in the ratio 3 : 1

Then by Section formula, coordinates of P are
\(\left(\frac{3 \times 9+1 \times 5}{3+1}, \frac{3 \times 6+1 \times(-2)}{3+1}\right)\)
i.e. P(8, 4)

Question 3.
The vertices of a quad. PMQS are P(0, 0), M(3, 2), Q(7, 7) and S(4, 5). Show that PMQS is a parallelogram.
Solution:
Here PM = \(\sqrt{(3-0)^2+(2-0)^2}\) = \(\sqrt{13}\)
MQ = \(\sqrt{(7-3)^2+(7-2)^2}\)
= \(\sqrt{16+25}\) = \(\sqrt{41}\)
SQ = \(\sqrt{(7-4)^2+(7-5)^2}\)
= \(\sqrt{9+4}\) = \(\sqrt{13}\)

PS = \(\sqrt{(4-0)^2+(5-0)^2}\)
= \(\sqrt{16+25}\) = \(\sqrt{41}\)
∴ PM = SQ and PS = MQ
Thus opposite sides are equal. Mid-point of diagonal PQ be
\(\left(\frac{7+0}{2}, \frac{7+0}{2}\right)\) i.e. \(\left(\frac{7}{2}, \frac{7}{2}\right)\)
and Mid-point of diagonal MQ be
\(\left(\frac{3+4}{2}, \frac{2+5}{2}\right)\) i.e. \(\left(\frac{7}{2}, \frac{7}{2}\right)\)
Thus diagonals bisect each other.
Hence the quadrilateral PMQS is a parallelogram.
Also slope of PM = \(\frac{2-0}{3-0}\) = \(\frac{2}{3}\)
slope of SQ = \(\frac{7-5}{7-4}\) = \(\frac{2}{3}\)
∴ PM || SQ
Slope of MQ = \(\frac{7-2}{7-3}\) = \(\frac{5}{4}\)
and slope of PS = \(\frac{5-0}{4-0}\) = \(\frac{5}{4}\)
∴ MQ || PS

Question 4.
P, Q and R are three collinear points. P and Q are (3, 4) and (7, 7) respectively, and PR = 10 units. Find the coordinates of R.
Solution:
Let the coordinates of point R be (h, k).
Since P, Q and R are collinear
∴ area of △PQR = 0

⇒ \(\frac{1}{2}\) | (21 – 28) + (7k – 7h) + (4h – 3k) | = 0
⇒ – 7 – 3h – 4k = 0
⇒ 3h – 4k + 7 = 0 …(1)
Also PR = 10
⇒ PR² = 100
(h – 3)² + (k – 4)² = 100 …(2)
From (1) and (2); we have
\(\left(\frac{4 k-7}{3}-3\right)^2\) + (k – 4)² = 100
⇒ \(\frac{(4 k-16)^2}{9}\) + (k – 4)² = 100
⇒ \(\frac { 16 }{ 9 }\) (k – 4)² + (k – 4)² = 100
⇒ (k – 4)² × \(\frac { 25 }{ 9 }\) = 100
⇒ (k – 4)² = \(\frac{100 \times 9}{25}\) = 36
⇒ k – 4 = ± 6 ⇒ k = 4 ± 6 ⇒ k = 10, – 2
∴ from (1); h = \(\frac{4 \times 10-7}{3}\) = 11
and h = \(\frac{4(-2)-7}{3}\) = – 5
Hence the coordinates of point P are (11, 10) or (- 5, – 2).

Question 5.
The coordinates of the vertices of a triangle are (4, – 3), (- 5, 2) and (x, y). If the centroid of the triangle is at the origin, show that x = y = 1.
Solution:
Given vertices of triangle are A(4, – 3), B(- 5, 2) and C(x, y)
Thus centroid of $\triangle △ABC be given by
\(\left(\frac{4-5+x}{3}, \frac{-3+2+y}{3}\right)\) i.e. \(\left(\frac{x-1}{3}, \frac{y-1}{3}\right)\)
Also given centroid of △ABC be (0, 0).
Thus \(\frac{x-1}{3}\) = 0 ⇒ x = 1
and \(\frac{y-1}{3}\) = 0 ⇒ y = 1
∴ x = y = 1

Question 6.
Find the equation of the line passing through the point (- 4, – 5) and perpendicular to the line joining the points (1, 2) and (5, 6).
Solution:
Slope of line joining (1, 2) and (5, 6)

∴ slope of required line = – \(\frac { 1 }{ m }\) = –\(\frac { 1 }{ 1 }\) = -1
[since required line is ⊥ to line joining (1, 2) and (5, 6)]
Thus by using one point form, eqn. of line through the point (- 4, – 5) and having slope -1 is given by
y + 5 = – 1(x + 4)
| ∵ y – y₁ = m(x – x₁)
⇒ x + y + 9 = 0
which is the required line.

Question 7.
Find the equation of the straight line which passes through the point (3, 4) and has intercepts on the axes such that their sum is 14 .
Solution:
Let the required eqn. of line in intercept form be given by
\(\frac { x }{ a }\) + \(\frac { y }{ b }\) = 1 …(1)
Also a + b = 14 …(2)
eqn. (1) passes through the point (3, 4).
∴ \(\frac { 3 }{ a }\) + \(\frac { 4 }{ b }\) = 1 ⇒ 3b + 4a = ab
⇒ 3b + 4 (14 – b) = b (14 – b) [using (2)]
⇒ b² – 15b + 56 = 0
⇒ (b – 8) (b – 7) = 0
⇒ b = 7, 8
When b = 7 ∴ from (2); a = 7
When b = 8 ∴ from (2); a = 6
∴ from (1); eqns. of straight lines are given
by \(\frac { x }{ 7 }\) + \(\frac { y }{ 7 }\) = 1 ⇒ x + y = 7
and \(\frac { x }{ 6 }\) + \(\frac { y }{ 8 }\) = 1 ⇒ 4x + 3y = 24

Question 8.
Reduce the equation of the straight line 3x + 4y + 14 = 0 to normal form and find the perpendicular distance of the line from the origin.
Solution:
Given eqn. of line can be written as
3x + 4y = – 15 ⇒ – 3x – 4y = – 15
⇒ \(\frac{-3}{\sqrt{(-3)^2+(-4)^2}} x\) – \(\frac{4}{\sqrt{(-3)^2+(-4)^2}} y\) = \(\frac{15}{\sqrt{(3)^2+(-4)^2}}\)
⇒ –\(\frac { 3 }{ 5 }\)x – \(\frac { 4 }{ 5 }\)y = 3 …(1)
On comparing eqn. (1) with
x cos α + y sin α = p
we have –\(\frac { 3 }{ 5 }\) = cos α
and –\(\frac { 4 }{ 5 }\) = sin α and p = 3
On dividing; tan α = \(-\frac{-\frac{4}{5}}{-\frac{3}{5}}\) = \(\frac { 4 }{ 3 }\)
⇒ α = π + tan^-1
[∵ α lies in IIIrd quadrant]
Thus eqn. (1) reduces to ;
x cos α + y sin α = p
where p = 3 = ⊥ distance of the line from (0, 0).

Question 9.
Find the equation of the straight line which passes through the point of intersection of the straight lines x + y = 8 and 3x – 2y + 1 = 0 and is parallel to the straight line joining the points (3, 4) and (5, 6).
Solution:
Given eqns. of lines are
x + y – 8 = 0 …(1)
3x – 2y + 1 = 0 …(2)
Thus the eqn. of straight line which passes through the point of intersection of lines (1) and (2) is given by
x + y – 8 + k(3x – 2y + 1) = 0
⇒ (1 + 3k) x + (1 – 2k) y – 8 + k = 0
Thus slope of line (3) =\(\frac{-(1+3 k)}{1-2 k}\) = \(\frac{1+3 k}{2 k-1}\)
Also, slope of line joining the points (3, 4) and (5, 0) = \(\frac { 6-4 }{ 5-3 }\) = 1
Since line (3) is parallel to line joining the points (3, 4) and (5, 6). Thus their slopes are equal.
∴ \(\frac{1+3 k}{2 k-1}\) = 1 ⇒ 1 + 3k = 2k – 1
⇒ k = – 2
putting the value of k in eqn. (3) ; we have
– 5x + 5y – 10 = 0
⇒ x – y + 2 = 0
which is the required eqn.

Question 10.
Find the equation of the straight line which passes through the point of intersection of the straight lines 3x – 4y + 1 = 0 and 5x + y – 1 = 0 and cuts off equal intercepts from the axes.
Solution:
Given eqns. of lines are
3x – 4y + 1 = 0 …(1)
5x + y – 1 = 0 …(2)
Now the eqn. of line passes through the line of intersection of lines (1) and (2) be given by
3x – 4y + 1 + k(5x + y – 1) = 0 …(3)
⇒ (3 + 5k) x +( 4 + k) y + 1 – k = 0
⇒ (3 + 5 k) x + (k – 4) y = k – 1
⇒ \(\frac{x}{\frac{k-1}{3+5 k}}\) + \(\frac{y}{\frac{k-1}{k-4}}\) = 1
which is of intercept form having intercepts on x-axis and y-axis are \(\frac{k-1}{3+5 k}\) and \(\frac{k-1}{k-4}\).
Since it is given that both intercepts are of equal length.
∴ \(\frac{k-1}{3+5 k}\) = \(\frac{k-1}{k-4}\)
But k ≠ 1
∴ 3 + 5k = k – 4 ⇒ 4k = – 7
⇒ k = –\(\frac{7}{4}\)
putting the value of k in eqn. (3) ; we have
3x – 4y + 1 – \(\frac{7}{4}\) (5x + y – 1) = 0
⇒ – 23x – 23y + 11 = 0
⇒ 23x + 23y – 11 = 0
which is the required line.

Question 11.
Find the locus of a point such that the line segments having end points (2, 0) and (- 2, 0) subtend a right angle at that point.
Solution:
Let P(h, k) be any point on the locus.
Then by given condition, △PAB be an right angled triangle.

∴ PA² + PB² + AB²
⇒ (h – 2)² + (k – 0)² + (h + 2)² + (k – 0)² = (- 2 – 2)² + (0 – 0)²
⇒ h² – 4h + 4 + k² + h² + 4h + 4 + k² = 16
⇒ 2h² + 2k² – 8 = 0 ⇒ h² + k² = 4
Thus the required locus of P(h, k) be given by
by x² + y² = 4

Question 12.
Find the coordinates of the orthocentre of the triangle whose vertices are (1, 2), (2, 3) and (4, 3).
Solution:
Let the given vertices of triangle are A(1, 2), B(2, 3) and C(4, 3)
∴ slope of line BC = \(\frac { 3-3 }{ 4-2 }\) = 0
Thus, slope of altitude AL = ∞ [∵ AL ⊥ BC]

Thus by using one point form, eqn. of line AL be given by
x – 1 = 0 …(1)
| ∵ y – y₁ = m (x – x₁)
and slope of AC = \(\frac{2-3}{1-4}\) = \(\frac{-1}{-3}\) = \(\frac{1}{3}\)
Also, line BM ⊥ AC
∴ slope of line BM = – 3 [∵ m₁ m₂ = – 1]
slope of line AB = \(\frac{3-2}{2-1}\) = 1
∴ slope of line CN = –\(\frac{1}{1}\) = – 1 [∵ CN ⊥ AB]
Thus, eqn. of line through the point B(2, 3) and having slope -3 is given by
y – 3 = – 3(x – 2)
⇒ 3x + y – 9 = 0 …(2)
The eqn. of line passing through the point C(4, 3) and having slope -1 be given by
y – 3 = – 1(x – 4)
⇒ x + y – 7 = 0 …(3)
We know that, the orthocentre is the point of intersection of all three altitudes of the triangle.
lines (1) and (2) intersects at (1, 6)
Clearly the point (1, 6) lies on eqn. (3)
Thus (1, 6) be the point of concurrence of all three altitudes ∴(1, 6) be the required orthocentre of △ABC.

Question 13.
Find the equation of the bisector of the acute angle between the lines
3x – 4y + 7 = 0 and 12x + 5y – 2 = 0 .
Solution:
Given eqns. of lines can be written as (with positive constant terms)
3x – 4y + 7 = 0 …(1)
and – 12x – 5y + 2 = 0
Thus eqns. of angular bisectors of lines (1) and (2) be given by

Taking +ve sign ;
13(3x – 4y + 7) = 5(-12x – 5y + 2)
⇒ 99x – 27y – 81 = 0
⇒ 11x – 3y – 9 = 0 …(3)
Taking -ve sign ;
13(3x – 4y + 7) = – 5(- 12x – 5y + 2)
⇒ 21x + 77y – 101 = 0 …(4)
out of eqn. (3) and (4) that bisector will be the bisector of the acute angle which makes an angle θ < 45° with any of the lines (1) and (2).
Let us take the bisector 11x – 3y – 9 = 0 and line (1) i.e. 3x – 4y + 7 = 0
∴ slope of line (3) = \(\frac{-11}{-3}\) = \(\frac{11}{3}\)
slope of line (1) = \(\frac{-3}{-4}\) = \(\frac{3}{4}\)
Let θ be the angle between line (3) and line (1)

Thus 11x – 3y + 9 = 0 be the required eqn. of the bisector of acute angle between the given lines.