Very Short Answer on The Straight Line Class 11 OP Malhotra Exe-16K ISC Maths Solutions Ch-16. In this article you would learn to solve hard problems easily on The Straight Line. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
The Straight Line Class 11 OP Malhotra Very Short Answer ISC Math Solutions Ch-16
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-16 | The Straight Line |
| Writer | OP Malhotra |
| Exe-16(K) | Very Short Answer Type Questions. |
Very Short Answer on The Straight Line
OP Malhotra ISC Class 11 Maths Solutions
Que-1: Find a point on the x-axis which is equidistant from the points (7, 6) and (3, 4).
Sol: Let the point on x-axis be (x, 0).
Distance from (7,6) = √[(x−7)² + 36]
Distance from (3,4) = √[(x−3)² + 16]
Equating both:
(x−7)² + 36 = (x−3)² + 16
⇒ x² −14x +49 +36 = x² −6x +9 +16
⇒ −14x +85 = −6x +25
⇒ −8x = −60 ⇒ x = 7.5
Hence, required point is (7.5, 0).
Que-2: Find the slope of the line passing through the origin and midpoint of A(0, −4) and B(8, 0).
Sol: Midpoint = ((0+8)/2, (−4+0)/2) = (4, −2)
Slope = (−2 − 0)/(4 − 0) = −2/4 = −1/2.
Que-3: If A(4,4), B(3,5) and C(−1,−1) are vertices of a triangle, show that AB ⟂ AC.
Sol: Slope AB = (5−4)/(3−4) = 1/(−1) = −1
Slope AC = (−1−4)/(−1−4) = (−5)/(−5) = 1
Product = (−1)(1) = −1
⇒ AB ⟂ AC.
Que-4: Find coordinates of the point dividing the line joining (5, −2) and (9, 6) in ratio 3:1.
Sol: Using section formula:
x = (3×9 + 1×5)/(3+1)
= (27+5)/4 = 8
y = (3×6 + 1×(−2))/4
= (18−2)/4 = 4
Required point = (8, 4).
Que-5: The vertices of triangle are (4, −3), (−5, 2) and (x, y). If centroid is at origin, find (x, y).
Sol: Centroid = (x₁+x₂+x₃)/3 , (y₁+y₂+y₃)/3 = (0,0)
So, 4 −5 + x = 0 ⇒ x = 1
−3 + 2 + y = 0 ⇒ y = 1
Hence, (x, y) = (1, 1).
Que-6: Find value of x such that points (x, −1), (2, 1) and (4, 5) are collinear.
Sol: Slope (2,1) & (4,5) = (5−1)/(4−2) = 4/2 = 2
Slope (x,−1) & (2,1) = (1+1)/(2−x) = 2/(2−x)
Equating: 2/(2−x) = 2
⇒ 2 = 2(2−x)
⇒ 2 = 4 −2x
⇒ 2x = 2
⇒ x = 1.
Que-7: Find angle between x-axis and line joining (3, −1) and (4, −2).
Sol: Slope = (−2+1)/(4−3) = −1
Angle θ = tan⁻¹(−1) = −45°
Angle with x-axis = 45°.
Que-8: Find equation of a straight line whose perpendicular distance from origin is 4 units and makes angle 120° with positive x-axis.
Sol: Using normal form:
x cosθ + y sinθ = p
Here θ = 120°, p = 4
cos120° = −1/2, sin120° = √3/2
So equation:
x(−1/2) + y(√3/2) = 4
⇒ −x + √3y = 8
Required equation: −x + √3y = 8.
Que-9: Reduce the equation 3x + 4y + 15 = 0 to normal form and find perpendicular distance from origin.
Sol: Divide by √(3²+4²)=5:
(3/5)x + (4/5)y + 3 = 0
⇒ (3/5)x + (4/5)y = -3
Normal form: x cosθ + y sinθ = p
Here p = 3 (distance from origin).
Hence perpendicular distance = 3 units.
Que-10: Find equation of line passing through (1,2) and making angle 30° with y-axis.
Sol: Angle with x-axis = 60°
Slope m = tan60° = √3
Using point-slope form:
y − 2 = √3(x − 1)
Que-11: Find equation of line through (1, −2) cutting equal intercepts.
Sol: Intercept form: x/a + y/a =1 ⇒ x+y=a
Put (1,−2): 1−2=a ⇒ a=−1
Equation: x + y = −1
x + y + 1 = 0
Que-12: Check if lines joining (3, −4),(−2,6) and (−3,6),(9,−18) are perpendicular.
Sol: Slope₁ = (6+4)/(−2−3)=10/−5=−2
Slope₂ = (−18−6)/(9+3)=−24/12=−2
Product ≠ −1
⇒ Not perpendicular.
It is parallel.
Que-13: Find angle between lines joining (0,0)-(2,3) and (2,−2)-(3,5).
Sol: m₁ = 3/2, m₂ = 7
tanθ = |(m₂−m₁)/(1+m₁m₂)|
= (7−3/2)/(1+21/2) = (11/2)/(23/2)
= 11/23
Que-14: Find equation of line through (−1, −2) and (−5, 2).
Sol: Slope = (2+2)/(−5+1)
= 4/−4 = −1
y + 2 = −1(x +1)
⇒ y = −x −3
x + y + 3 = 0.
Que-15: Find equation through intersection of 3x+4y=7 and x−y+2=0 with slope 3.
Sol: Solve: x−y+2 = 0
⇒ y = x + 2
Substitute: 3x+4(x+2) = 7
⇒ 7x + 8 = 7
⇒ x = −1, y = 1
Equation: y−1 = 3(x+1)
Que-16: Find distance of point (−1,1) from line 12(x+6)=5(y−2).
Sol: 12x−5y+82=0
Distance = |12(−1)−5(1)+82| / √(144+25)
= 65/13 = 5
Que-17: Find distance between parallel lines 15x+8y−34=0 and 15x+8y+31=0.
Sol: Distance = |31+34| / √(225+64)
= 65/17
Que-18: Find equation of line perpendicular to x/4 + y/6 =1 at y-axis.
Sol: Given slope = −3/2
⇒ perpendicular slope = 2/3
Point on y-axis: (0,6)
Equation: y−6 = (2/3)x
3y – 18 = 2x
2x – 3y + 18 = 0
Que-19: Find p such that lines 3x+y−2=0, px+2y−3=0 and 2x−y−3=0 are concurrent.
Sol: Solve first & third:
3x+y=2 and 2x−y=3
⇒ x = 1, y = −1
Put in second: p(1)+2(−1)−3 = 0
⇒ p − 2 − 3 = 0
⇒ p = 5.
Que-20: Find equation of angle bisector of 3x+4y+2=0 and 5x−12y−6=0.
Sol: Angle bisector formula:
(3x + 4y + 2)/√(3² + 4²) = (5x − 12y − 6)/√(5² + 12²)
⇒ (3x + 4y + 2)/5 = (5x − 12y − 6)/13
Cross multiply:
13(3x + 4y + 2) = 5(5x − 12y − 6)
⇒ 39x + 52y + 26 = 25x − 60y − 30
⇒ 14x + 112y + 56 = 0
⇒ x + 8y + 4 = 0
Other bisector:
13(3x + 4y + 2) = −5(5x − 12y − 6)
⇒ 39x + 52y + 26 = −25x + 60y + 30
⇒ 64x − 8y − 4 = 0
⇒ x − 2y − 1 = 0
Que-21: Find angle between √3x−y=0 and x−√3y+1=0.
Sol: Slopes m₁=√3, m₂=1/√3
tanθ = (m₁−m₂)/(1+m₁m₂)
= (√3−1/√3)/(1+1)
= 1/√3
⇒ θ = 30°
Que-22: Show that tanθ between x/a + y/b =1 and x/a − y/b =1 is 2ab/(a²−b²).
Sol: Slopes m₁ = −b/a and m₂ = b/a
tanθ = (m₂−m₁)/(1+m₁m₂)
= (2b/a)/(1−b²/a²)
= 2ab/(a²−b²)
–: End The Straight Line Class 11 OP Malhotra Exe-16K ISC Maths Ch-16 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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