11th Grade > Mathematics
STRAIGHT LINES MCQs
Total Questions : 30
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Answer: Option C. ->
(2, - 3)
:
C
:
C
Putting x = x' + h, y = y' + k, the given equation transforms to
x′2+y′2+x′(2h−4)+y′(2k+6)+h2+k2−7=0
To eliminate linear terms, we should have
2h - 4 = 0, 2k + 6 = 0 ⇒ h = 2, k = -3
i.e., (h, k) = (2, -3).
Answer: Option A. ->
4,83
:
A
Sol: Let A ≡ (0, 0), B ≡ (8, 0) and C ≡ (4, 6).
Slope of BC = 6−04−0=32
Equation of the line through A(0, 0) and perpendicular to BC is
(y – 0) = 23 (x – 0) i.e. 2x – 3y = 0 …… (1)
Slope of CA =6−04−0=32
Equation of the line through B(8, 0) and perpendicular to CA is
(y – 0) = −23 (x – 8) i.e., 2x + 3y = 16 …… (2)
Solving (1) and (2), the orthocenter is 4,83
:
A
Sol: Let A ≡ (0, 0), B ≡ (8, 0) and C ≡ (4, 6).
Slope of BC = 6−04−0=32
Equation of the line through A(0, 0) and perpendicular to BC is
(y – 0) = 23 (x – 0) i.e. 2x – 3y = 0 …… (1)
Slope of CA =6−04−0=32
Equation of the line through B(8, 0) and perpendicular to CA is
(y – 0) = −23 (x – 8) i.e., 2x + 3y = 16 …… (2)
Solving (1) and (2), the orthocenter is 4,83
Answer: Option C. ->
3x + y + 7 = 0 or x – 3y – 31 = 0
:
C
Any line through (1, - 10) is given by y + 10 = m(x - 1)
Since it makes equal angle ′α′ with the given lines 7x – y + 3 = 0 and x + y – 3 = 0, therefore
tanα=m−71+7m=m+11+m(−1)⇒m=13 or −3
Hence the two possible equations of third side are 3x + y + 7 = 0 and x - 3y - 31 = 0.
:
C
Any line through (1, - 10) is given by y + 10 = m(x - 1)
Since it makes equal angle ′α′ with the given lines 7x – y + 3 = 0 and x + y – 3 = 0, therefore
tanα=m−71+7m=m+11+m(−1)⇒m=13 or −3
Hence the two possible equations of third side are 3x + y + 7 = 0 and x - 3y - 31 = 0.
Answer: Option D. ->
2x – 9y + 7 = 0
:
D
S = midpoint of QR = (6+72,−1+32)=(132,1)∴slope of PS=2−12−132=−29∴The required equation is y+1=−29(x−1)i.e.,2x+9y+7=0
:
D
S = midpoint of QR = (6+72,−1+32)=(132,1)∴slope of PS=2−12−132=−29∴The required equation is y+1=−29(x−1)i.e.,2x+9y+7=0
Answer: Option C. ->
3¯x =2α−u and 3¯y =2β+v
:
C
We know that, the centroid of a triangle divides the segment joining the orthocentre and circumcentre internally in the ratio 2 : 1. Therefore,
¯x=2α+u2+1and¯y=2β+v2+1
⇒3¯x=2α+uand¯y=2β+v
:
C
We know that, the centroid of a triangle divides the segment joining the orthocentre and circumcentre internally in the ratio 2 : 1. Therefore,
¯x=2α+u2+1and¯y=2β+v2+1
⇒3¯x=2α+uand¯y=2β+v
Answer: Option C. ->
112
:
C
:
C
Distance between parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is |c1−c2|√a2+b2
Given lines are 8x+6y+5 = 0 and 4x + 3y - 25 = 0
⇒ 8x + 6y - 50 = 0
∴ Required distance = 5+50√82+62=5510=112
Answer: Option C. ->
tan−1√335
:
C
:
C
tanθ=2√h2−aba+b
θ=tan−12√494−45=tan−1√335.
Answer: Option C. ->
60∘
:
C
θ=tan−1(2√h2−aba+b)=tan−1(√4h2−4aba+b)
:
C
θ=tan−1(2√h2−aba+b)=tan−1(√4h2−4aba+b)
= tan−1(√3a2+3b2+10ab−4aba+b)=60∘.