In a triangle ABC- a point P is chosen on side AB such that AP - PB -1- 4 and a point Q is chosen on side BC such that CQ - QB -1- 3- Line segment C P and A Q intersect at M- If the ratio M C-P C is expressed as a rational number in the lowest term as a-h- then b- a equals

Vector equation of AQ is r_1=a+λ(a 3c4) nbsp; and vector equation of CP is r_2=c+μ nbsp;(c 4a5) nbsp; r_1=r_2 gives nbsp; a+λ(a 3c4)=c+μ nbsp;(c 4a5) ...

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Byju's Answer

Standard XII

Mathematics

Vector Equation for Straight Line

In a triangle...

Question

In a triangle $A B C,$ a point $P$ is chosen on side $- - \to A B$ such that $A P : P B = 1 : 4$ and a point $Q$ is chosen on side $- - \to B C$ such that $C Q : Q B = 1 : 3.$ Line segment $- - \to C P$ and $- - \to A Q$ intersect at $M .$ If the ratio $\frac{M C}{P C}$ is expressed as a rational number in the lowest term as $\frac{a}{b},$ then $b - a$ equals

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Solution

Vector equation of $A Q$ is $_{1} = \to a + λ (\to a - \frac{3 \to c}{4})$
and vector equation of $C P$ is
$_{2} = \to c + μ (\to c - \frac{4 \to a}{5})$
$_{1} =_{2}$ gives
$\to a + λ (\to a - \frac{3 \to c}{4}) = \to c + μ (\to c - \frac{4 \to a}{5})$
On comparing both sides, we get
$1 + λ + \frac{4 μ}{5} = 0$
and $1 + μ + \frac{3 λ}{4} = 0$
Solving, $λ = - \frac{1}{2}, μ = - \frac{5}{8}$

$∴$ Position vector of $M$ are $\to a - \frac{1}{2} \to a + \frac{3 \to c}{8} = \frac{4 \to a + 3 \to c}{8}$
(putting value of $λ$ in $r_{1}$ )

$| - - \to M C | = ∣ ∣ ∣ ∣ \to c - \frac{4 \to a + 3 \to c}{8} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \frac{5 \to c - 4 \to a}{8} ∣ ∣ ∣ ∣$
and $| - - \to P C | = ∣ ∣ ∣ ∣ \to c - \frac{4 \to a}{5} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \frac{5 \to c - 4 \to a}{5} ∣ ∣ ∣ ∣$

$∴ \frac{M C}{P C} = \frac{5}{8}$
$\Rightarrow a = 5$ and $b = 8$
$∴ b - a = 3$

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Similar questions

Q. In a triangle

A B C,

a point

P

is chosen on side

- - \to A B

such that

A P : P B = 1 : 4

and a point

Q

is chosen on side

- - \to B C

such that

C Q : Q B = 1 : 3.

Line segment

- - \to C P

and

- - \to A Q

intersect at

M .

If the ratio

\frac{M C}{P C}

is expressed as a rational number in the lowest term as

\frac{a}{b},

then

b - a

equals

Q. In a

△ A B C

a point

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is taken such that

\frac{A P}{B P} = \frac{1}{3}

and a point

Q

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such that

\frac{C Q}{B Q} = \frac{3}{1} .

R

is the point of intersection of the lines

A Q

and

C P

. The area of

△ B R C

1

unit and area of

△ A B C

\frac{a}{b}, a and b

is coprime, then

a + b

Q. Let ABC be a triangle and M be a point on side AC closer to vertex C than A. Let N be a point on side AB such that MN is parallel to BC and let P be a point on side BC such that MP is parallel to AB. If the area of the quadrilateral BNMP is equal to

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Q. In

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a point

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\frac{A P}{B P} = \frac{1}{3}

and a point

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\frac{C Q}{B Q} = \frac{3}{1} .

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In a triangle ABC, a point P is chosen on side −−→AB such that AP:PB=1:4 and a point Q is chosen on side −−→BC such that CQ:QB=1:3. Line segment −−→CP and −−→AQ intersect at M. If the ratio MCPC is expressed as a rational number in the lowest term as ab, then b−a equals