Question

The line segment AB meets the coordinate axes in points A and B. If point P (3,6) divides
AB in the ratio 2:3, then find the points A and B.​

Answer 1

Answer:

Let A(α,0),B(0,β)

By section formula,

−5=

5

3α

⇒α=

3

−25

and 2=

5

2β

⇒β=5

⇒ By intercept form, equation of a line

α

x

+

β

y

=1

⇒

−25

3(x)

+

5

y

=1

⇒−3x+5y=25

Answer 2

$\huge\sf\pink{Answer}$

☞ The points are, A(5,0) & B(0,15)

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$\huge\sf\blue{Given}$

✭ Line segment AB is divided by point P(3,6) in the ratio 2:3

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$\huge\sf\gray{To \:Find}$

◈ The points AB?

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$\huge\sf\purple{Steps}$

$\large\underline{\underline{\sf Concept}}$

So here we shall use the section formula to find the coordinates of A & B. As we are given that the points A & B meets the axes,i.e point A(x,0) & point B(0,y).You may refer to the attachment to get a clear idea of the section formula

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We shall here use the section formula, that is,

$\underline{\boxed{\sf x = \dfrac{m_1 x_2+m_2x_1}{m+n}}}$

Similarly,

$\underline{\boxed{\sf y = \dfrac{m_1y_2+m_2y_1}{m+n}}}$

So here we see that,

◕ $\sf m_1 \ \& \ m_1 = 2,3$

◕ $\sf x_1 \ \& \ x_2 = x,0$

◕ $\sf y_1 \ \& \ y_2 = 0,y$

So we shall first try to find the value of x,

➝ $\sf x = \dfrac{m_1 x_2+m_2x_1}{m+n}$

➝ $\sf 3 = \dfrac{2(0)+3(x)}{2+3}$

➝ $\sf 3 = \dfrac{0+3x}{5}$

➝ $\sf 3\times 5 = 3x$

➝ $\sf 15 = 3x$

➝ $\sf \dfrac{15}{3} = x$

➝ $\sf \red{x=5}$

Now it's the time to find the value of y,

➳ $\sf y = \dfrac{m_1y_2+m_2y_1}{m+n}$

➳ $\sf 6 = \dfrac{2(y)+0}{5}$

➳ $\sf 6\times 5 = 2y+0$

➳ $\sf 30 = 2y$

➳ $\sf \dfrac{30}{2} = y$

➳ $\sf \orange{y=15}$

$\sf \bullet\:\: A = (5,0)$

$\sf \bullet\:\: B = (0,15)$

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