Question

The point A (4,7), B (p,3) and C (7,3) are the vertices of right angled triangle, right angled at point B, find the values of p

Answer 1

Hint: To solve this question, first we will put Pythagoras theorem, then we will use coordinates of points to obtain the length of the sides of the triangle using the formula ‘if X (a,b) and Y (a’,b’), then $X{Y^2} = {(a' - a)^2} + {(b' - b)^2}$, thus we will get a equation containing variable p. Solving that equation by middle term factorization method or by using quadratic formula, we will get the value of p.

Complete step-by-step answer:
According to the question, the points A (4,7), B (p,3) and C (7,3) are the vertices of the right angled triangle.
And the triangle is right angled at point B.
Hence, $m\angle B = 90^\circ $
               

That means AC is the hypotenuse of the right angled triangle and AB and BC are perpendicular and base. (As the side that is in front of the right angle is hypotenuse.
According to the Pythagoras theorem, in the right angled triangle, the square of the length of the hypotenuse is the sum of the square of the length of the base and the square of the length of the perpendicular.
Hence, $A{C^2} = A{B^2} + B{C^2}$……….(1)
Using coordinates to get the lengths of the sides and putting them in above equation we get,
${(7 - 4)^2} + {(3 - 7)^2} = {(p - 4)^2} + {(3 - 7)^2} + {(p - 7)^2} + {(3 - 3)^2}$
Simplifying the above equation we get,
${3^2} + {( - 4)^2} = {(p - 4)^2} + {( - 4)^2} + {(p - 7)^2} + 0$
$ \Rightarrow 9 + 16 = {p^2} + 16 - 8p + 16 + {p^2} + 49 - 14p$
 Calculating the above equation and putting all the terms into one side and taking them as equal to 0 we get,
$2{p^2} - 22p + 56 = 0$
Dividing two on the both side of the above equation we get,
${p^2} - 11p + 28 = 0$
Doing middle term factorization,
${p^2} - 7p - 4p + 28 = 0$
Taking p common from first two terms and -4 from second two terms we get,
$p(p - 7) - 4(p - 7) = 0$
$ \Rightarrow (p - 7)(p - 4) = 0$
Taking, $p - 7 = 0$
We got p=7, which is not as because B and C will be the same.
And again taking, $p - 4 = 0$
We got p=4
Hence, the values of $p = 4$

Note: Pythagoras theorem states that in the right angled triangle, the square of the length of the hypotenuse is the sum of the square of the length of the base and the square of the length of the perpendicular.
If X (a, b) and Y (a’, b’), then $X{Y^2} = {(a' - a)^2} + {(b' - b)^2}$. This is the formula of obtaining length from the coordinates of its end points. You should remember all the formulas of coordinate geometry.
To solve the quadratic equation we use two methods. One is quadratic formula of obtaining roots i.e. $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and another is middle term factorization in which we split coefficient of middle term in such a way that its product is equal to the last term.
You should practice more and more quadratic equations using these two methods.