Question

With the help of distance formula, show that the points A (4,2), B(7,5), and C(9,7) do not form a triangle.

Answer 1

$\underline{ { \colorbox{yellow}{Solution.}}}$

$\rm We \: have,$

$\: \: \: \: \: \: A \implies( 4, 2 )$

$\: \: \: \: \: B \implies( 7, 5 )$

$\: \: \: \: \: \: C \implies( 9, 7 )$

$\bf \red{By \: Using \: Distant \: Formula}$

$\: \: \: \: \sf \purple{ \: \: AB }= \: \: \sqrt[]{(7 - 4) {}^{2} { + (5 - 2) {}^{} }^{2} }$

$\sf \: \: \: \: \: \: \: = \: \: \: \sqrt[]{(3) {}^{2} } + (3) {}^{2}$

$\: \: \: \: \: \: \: \: \: \sf = \: \sqrt[]{9 + 9 \: } = \sqrt[]{13 \: }$

$\: \: \: \: \sf \purple{ BC} = \sqrt[]{(9 - 7) {}^{2} +( 7 - 5) {}^{2} }$

$\: \: \: \: \: \: \sf = \: \sqrt[]{4 + 4 \: } = \sqrt{8 \: }$

$\: \: \: \: \: = \sqrt{4 \times 2 \: } = 2 \sqrt{2 \: }$

$\rm and$

$\: \: \: \sf \purple{ CA} = \sqrt{(4 - 9) {}^{2} +( 2 - 7)} {}^{2}$

$\: \: \: \: \: \: \: \: \: = \sqrt{( - 5) {}^{2} + ( - 5) {}^{2} }$

$\: \: \: \: \: \: \: \: \sf = \sqrt{25 + 25 \: \: \: \: } \: \: \: \: = \sqrt{50}$

$\sf \sqrt{25 \times 2 \: } = 5 \sqrt{2 \: }$

$\sf \red{We \: find \: that}$

$\: \: \: \: \: \: \: \: \bf </strong><strong>AB</strong><strong> + </strong><strong>BC</strong><strong> = 3 \sqrt{2} + 2 \sqrt{2}$

$\: \: \: \: \: \sf = \: \: (3 + 2) \sqrt{2 \: }$

$\sf = 5 \sqrt{2 \: } = </strong><strong>CA</strong><strong>$

$\implies\tt \: </strong><strong>T</strong><strong>he \ points \: </strong><strong>A</strong><strong> \: </strong><strong>B</strong><strong> \: and \: </strong><strong>C</strong><strong> \: are \\ \: \: \: \: \: \: \: \: \: \: \: \: \ \tt colliniar.$

$\implies \tt \: </strong><strong>T</strong><strong>he \: points \: </strong><strong>A</strong><strong> \: </strong><strong>B </strong><strong>\</strong><strong>:</strong><strong> </strong><strong>and</strong><strong> \: </strong><strong>C</strong><strong> \: do \: not \: \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: form \: a \: triangle.$

Answer 2

AnswEr :

$\:\:\:\small\bullet$ These points are collinear.

$\:\:\:\small\bullet$ Points do not form a triangle.

$\rule{100}2$

ExplaNation :

$\underline{\bigstar\:\textsf{According \: to \: Question:}}$

$\:\:\:\bullet$ Three points are given :-

$\:\:\:\:\:\star$ A = (4,2)

$\:\:\:\:\:\star$ B = (7,5)

$\:\:\:\:\:\star$ C = (9,7)

$\:\:\:\bullet$ We use Distance formula to find sides(i.e. AB,BC and CA).

$\star{\boxed{\sf{Distance \: formula = \sqrt{(x_2- x_1)^2 + (y_2 - y_1)^2} }}}$

$\:\:\:\bullet$ Now; Let's solve

$\:\:\star$ Distance of AB -

$\normalsize\hookrightarrow\sf\sqrt{(7- 4)^2 + (5 - 2)^2} \\ \\ \normalsize\hookrightarrow\sf\sqrt{(3)^2 + (3)^2} \\ \\ \normalsize\hookrightarrow\sf\sqrt{(9+9)} \\ \\ \normalsize\hookrightarrow\sf\sqrt{18} \: = \: \sqrt{2 \times\ 3 \times\ 3} \\ \\ \normalsize\hookrightarrow\sf\ AB = 3\sqrt{2} \: units$

$\:\:\star$ Distance of BC -

$\normalsize\hookrightarrow\sf\sqrt{(9-7)^2 + (7-5)^2} \\ \\ \normalsize\hookrightarrow\sf\sqrt{(2)^2 + (2)^2} \\ \\ \normalsize\hookrightarrow\sf\sqrt{(4 + 4)} \\ \\ \normalsize\hookrightarrow\sf\sqrt{8} \: = \: \sqrt{2 \times\ 2 \times\ 2} \\ \\ \normalsize\hookrightarrow\sf\ BC \: = \: 2\sqrt{2} \: units$

$\:\:\star$ Distance of AC -

$\normalsize\hookrightarrow\sf\sqrt{(4-9)^2 + (2-7)^2} \\ \\ \normalsize\hookrightarrow\sf\sqrt{(-5)^2 + (-5)^2} \\ \\ \normalsize\hookrightarrow\sf\sqrt{ 25 + 25} \\ \\ \normalsize\hookrightarrow\sf\sqrt{50} \: = \: \sqrt{ 2 \times\ 5 \times\ 5} \\ \\ \normalsize\hookrightarrow\sf\ AC \: = \: 5\sqrt{2} \: units$

$\:\:\:\bullet$ Now; we know that a triangle is formed by points if Sum of two sides is greater than third side(i.e. AB + BC > AC), Let's check,by blocking the values in available data.

$\normalsize\hookrightarrow\sf\ AB + BC > AC$

$\normalsize\hookrightarrow\sf\ 3\sqrt{2} + 2\sqrt{2} > 5\sqrt{2} \\ \\ \normalsize\hookrightarrow\sf\ 5\sqrt{2} = 5\sqrt{2}$

$\:\:\:\bullet$ Here we can see,that sum of two points is equal to the third side (i.e. AB + BC = AC), so these points do not form a triangle.

$\Large\hookrightarrow{\underline{\boxed{\sf\green{Hence \: prove \:!!}}}}$