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Answers
EXPLANATION.
Ratio in which the line 2y - 3x = 8 divides the line segment joining the points (1, 4) and (-2,3).
As we know that,
Section formula for internal division.
\sf \implies \dfrac{mx_{2} + nx_{1}}{m + n} \ , \ \dfrac{my_{2} + ny_{1}}{m + n}⟹
m+n
mx
2
+nx
1
,
m+n
my
2
+ny
1
Using this formula in the equation, we get.
Let the line 2y - 3x = 8 divides the line segment in the ratio = k : 1.
⇒ x₁ = 1 and y₁ = 4.
⇒ x₂ = - 2 and y₂ = 3.
⇒ m = k and n = 1.
Put the values in the equation, we get.
\sf \implies \dfrac{k(-2) + 1(1)}{k + 1} \ , \ \dfrac{k(3) + 1(4)}{k + 1}⟹
k+1
k(−2)+1(1)
,
k+1
k(3)+1(4)
\sf \implies \dfrac{-2k + 1}{k + 1} \ , \ \dfrac{3k + 4}{k + 1}⟹
k+1
−2k+1
,
k+1
3k+4
Equation of line : 2y - 3x = 8.
Put the values of x and y in the equation of line, we get.
\sf \implies 2 \bigg[ \dfrac{3k + 4}{k + 1} \bigg] - 3 \bigg[ \dfrac{-2k + 1}{k + 1} \bigg] = 8⟹2[
k+1
3k+4
]−3[
k+1
−2k+1
]=8
\sf \implies \bigg[ \dfrac{6k + 8}{k + 1} \bigg] - \bigg[ \dfrac{- 6k + 3}{k + 1} \bigg] = 8⟹[
k+1
6k+8
]−[
k+1
−6k+3
]=8
\sf \implies (6k + 8) - (-6k + 3) = 8(k + 1)⟹(6k+8)−(−6k+3)=8(k+1)
\sf \implies 6k + 8 + 6k - 3 = 8k + 8⟹6k+8+6k−3=8k+8
\sf \implies 12k + 5 = 8k + 8⟹12k+5=8k+8
\sf \implies 12k - 8k = 8 - 5⟹12k−8k=8−5
\sf \implies 4k = 3⟹4k=3
\sf \implies k = \dfrac{3}{4}⟹k=
4
3
Ratio = 3 : 4.
EXPLANATION.
Ratio in which the line 2y - 3x = 8 divides the line segment joining the points (1, 4) and (-2,3).
As we know that,
Section formula for internal division.
\sf \implies \dfrac{mx_{2} + nx_{1}}{m + n} \ , \ \dfrac{my_{2} + ny_{1}}{m + n}⟹
m+n
mx
2
+nx
1
,
m+n
my
2
+ny
1
Using this formula in the equation, we get.
Let the line 2y - 3x = 8 divides the line segment in the ratio = k : 1.
⇒ x₁ = 1 and y₁ = 4.
⇒ x₂ = - 2 and y₂ = 3.
⇒ m = k and n = 1.
Put the values in the equation, we get.
\sf \implies \dfrac{k(-2) + 1(1)}{k + 1} \ , \ \dfrac{k(3) + 1(4)}{k + 1}⟹
k+1
k(−2)+1(1)
,
k+1
k(3)+1(4)
\sf \implies \dfrac{-2k + 1}{k + 1} \ , \ \dfrac{3k + 4}{k + 1}⟹
k+1
−2k+1
,
k+1
3k+4
Equation of line : 2y - 3x = 8.
Put the values of x and y in the equation of line, we get.
\sf \implies 2 \bigg[ \dfrac{3k + 4}{k + 1} \bigg] - 3 \bigg[ \dfrac{-2k + 1}{k + 1} \bigg] = 8⟹2[
k+1
3k+4
]−3[
k+1
−2k+1
]=8
\sf \implies \bigg[ \dfrac{6k + 8}{k + 1} \bigg] - \bigg[ \dfrac{- 6k + 3}{k + 1} \bigg] = 8⟹[
k+1
6k+8
]−[
k+1
−6k+3
]=8
\sf \implies (6k + 8) - (-6k + 3) = 8(k + 1)⟹(6k+8)−(−6k+3)=8(k+1)
\sf \implies 6k + 8 + 6k - 3 = 8k + 8⟹6k+8+6k−3=8k+8
\sf \implies 12k + 5 = 8k + 8⟹12k+5=8k+8
\sf \implies 12k - 8k = 8 - 5⟹12k−8k=8−5
\sf \implies 4k = 3⟹4k=3
\sf \implies k = \dfrac{3}{4}⟹k=
4
3
Ratio = 3 : 4.