Question

22) Find ratio in which the point P(-4, 6) divide the line segment joining the points
A(-6, 10) and B (3,- 8)​

Answer 1

❍ Let's say, that AB is the line segment and given points are A(–6, 10) and point B(3, –8) where point P(–4, 6) is intersecting the line in ratio of k: 1.

⠀

$\setlength{\unitlength}{14mm}\begin{picture}(7,5)(0,0)\thicklines\put(0,0){\line(1,0){5}}\put(5.1, - 0.3){\sf B}\put( - 0.2, - 0.3){\sf A}\put(5.2, 0){\sf (3,-8)}\put( - 0.7, 0){\sf (-6,10)}\put(2.3, 0.2){\sf P}\put(2.2, - 0.3){\sf (-4,6)}\put(5, 0){\circle*{0.1}}\put(2.4, 0){\circle*{0.1}}\put(0, 0){\circle*{0.1}}\put(1,0.2){\sf k}\put(3.5, 0.2){\sf 1}\end{picture}$⠀⠀

⠀

$\underline{\bf{\dag} \:\mathfrak{Using~Section~formula\: :}}$

$\bf{\dag}\;\boxed{\sf{ \: \: x = \bigg(\dfrac{m_{1} x_{2} + m_{2} x_{1}}{m_{1}+m_{2}}\bigg) \: \: }}$

⠀

• x₁ = – 6
• x₂ = 3
• x = – 4
• y = 6
• y₁ = 10
• y₂ = – 8
• m₁ = k
• m₂ = 1

Therefore,

⠀

$\dashrightarrow\sf -4 = \dfrac{k(3) + 1 (-6)}{k+1} \\\\\\\dashrightarrow\sf -4 = \dfrac{3k - 6}{k+1} \\\\\\\dashrightarrow\sf -4(k + 1) = 3k - 6 \\\\\\\dashrightarrow\sf -4k - 4 =3k - 6\\\\\\\dashrightarrow\sf -4k - 3k = -6+4\\\\\\\dashrightarrow\sf -7k = -2\\\\\\\dashrightarrow\sf k = \dfrac{\cancel{-}~2}{\cancel{-}~7}\\\\\\\dashrightarrow\sf k = \dfrac{2}{7}\\\\\\\dashrightarrow\underline{\boxed{\frak{\pmb{\pink{k:1 = 2:7}}}}}\;\bigstar$

⠀

$\therefore{\underline{\textsf{Hence, the ratio in which the given point dividing the line is \textbf{2: 7}.}}}$

Answer 2

Given:-

• Given point = P(-4, 6)

• Points of given line = A(-6, 10) and B (3,- 8)

To Find:-

• Ratio in which the point P(-4, 6) divide the line segment joining the points A(-6, 10) and B (3,- 8).

Formula Used:-

• Section Formula:

• $C=\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}$

Solution:-

According to the question,

Coordinates of Given line,

• A = (−6,10)

• B = (3,−8)

Now, there is a point C having the coordinates (−4,6) which divides the line segment AB.

Let us assume that point C divides the line segment AB in the ratio k:1, where k is a constant.

By section formula,

$C=\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}$ ____ {1}

Now, substituting

• $(x_1,y_1) = (−6,10)$

• $(x_2,y_2) = (3,−8)$

• m:n = k:1 in equation (1), we get,

$(-4,6)=\dfrac{3k-6}{k+1},\dfrac{10-8k}{k+1}$

Now, let us compare the x coordinates,

⇒$-4= \dfrac{3k-6}{k+1}$

⇒$-4k-4 =3k-6$

⇒$7k=2$

⇒$k=\dfrac{2}{7}$

Hence, the ratio k : 1 becomes $\dfrac{2}{7}:1.$

${\boxed{\red{k : 1 = 2 : 7}}}$

Therefore, the point C(−4,6) divides the line segment joining the points A(−6,10) and B(3,−8) in the ratio 2:7.

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