Video: CBSE Class X • Pack 4 • 2015 • Question 9

CBSE Class X • Pack 4 • 2015 • Question 9


Video Transcript

The points 𝐴: four, seven; 𝐵: 𝑝, three; and 𝐶: seven, three are the vertices of a right-angled triangle at 𝐵. Find the values of 𝑝.

We know that, for any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This is commonly written as 𝐴 squared plus 𝐵 squared equals 𝐶 squared, where 𝐶 is the length of the hypotenuse in the triangle. It’s also called the Pythagorean theorem.

Since we know this triangle is right-angled at 𝐵, this means that the sides 𝐴𝐵 and 𝐵𝐶 are the shorter sides of the triangle. This is because the hypotenuse always sits opposite the right angle. We can rewrite the Pythagorean theorem as 𝐴𝐵 squared plus 𝐵𝐶 squared equals 𝐴𝐶 squared.

We can use the distance formula to help us find the lengths of each of these sides. For the coordinates 𝑥 one, 𝑦 one and 𝑥 two, 𝑦 two, the distance between them is given by the formula shown. It’s the square root of the difference between the two 𝑥-coordinate squared plus the difference between the two 𝑦-coordinate squared. The distance between 𝐴𝐵 then is the square root of seven minus three all squared plus four minus 𝑝 all squared.

It’s important to note that this could be performed either way round. We could have subtracted seven from three and squared that and subtracted four from 𝑝 and squared that. Either method will give us the exact same answer.

Seven minus three all squared is four squared. And to square four minus 𝑝, it’s useful to write it as the product of two brackets. Four squared is 16. And to find the product of these two brackets, we first multiply four by four, which is 16. We then multiply the outer terms in the brackets. That’s four multiplied by negative 𝑝, which is negative four 𝑝. We then multiply the two inner terms, which is once again negative four 𝑝.

And finally, we multiply the last two terms in each bracket. Negative 𝑝 multiplied by negative 𝑝 is 𝑝 squared. 𝐴𝐵 is equal to the square root of 32 minus eight 𝑝 plus 𝑝 squared. For 𝐵 and 𝐶, the distance between these two points is three minus three all squared plus seven minus 𝑝 all squared. And three minus three is zero. Zero squared is zero. So we end up with the square root of seven minus 𝑝 all squared, which is just seven minus 𝑝.

The distance 𝐴𝐶 is the square root of three minus seven all squared plus seven minus four all squared. Three minus seven is negative four, and seven minus four is three. 𝐴𝐶 is the square root of 16 plus nine. 16 plus nine is 25. So since the square root of 25 is five, the length 𝐴𝐶 is simply five.

Let’s substitute each of these expressions into the Pythagorean formula. Notice that 𝐴𝐵 is the square root of 32 minus eight 𝑝 plus 𝑝 squared. If we square this expression, we just get 32 minus eight 𝑝 plus 𝑝 squared. 𝐵𝐶 squared becomes seven minus 𝑝 all squared, and 𝐴𝐶 squared is five squared.

Once again, let’s treat seven minus 𝑝 all squared as the product of two brackets. And when we expand this, we get 49 minus 14𝑝 plus 𝑝 squared. Our equation becomes 32 minus eight 𝑝 plus 𝑝 squared plus 49 minus 14𝑝 plus 𝑝 squared equals 25. And simplifying, we get two 𝑝 squared minus 22𝑝 plus 81 is equal to 25.

To solve a quadratic equation, we first need to make sure that the equation is equal to zero. So we’ll subtract 25 from both sides. That gives us two 𝑝 squared minus 22𝑝 plus 56 is equal to zero. Since each term is a multiple of two, we can divide everything by two to simplify this a little bit. And our equation becomes 𝑝 squared minus 11𝑝 plus 28 is equal to zero.

Now we need to factorise. Since the coefficient of 𝑝 squared is one, we know that the front part of each bracket is 𝑝. To find the constant part of each bracket, we’ll need to find two factors of 28 which have a sum of negative 11. The factors of 28 are one and 28, two and 14, and four and seven.

However, we know that the product of two negatives is also a positive. And since we’re trying to get a sum of negative 11, each factor of 28 should be negative. It’s negative one and negative 28, negative two and negative 14, and negative four and negative seven. Negative four and negative seven sum to negative 11. So our equation factorises to 𝑝 minus four multiplied by 𝑝 minus seven equals zero.

Now we know that we have two brackets whose product is zero. This means that at least one of these brackets must be equal to zero. Either 𝑝 minus four is equal to zero or 𝑝 minus seven is equal to zero.

Adding four to both sides in the first equation gives us that 𝑝 is equal to four. And adding seven in the second equation, we get 𝑝 is equal to seven. Notice though that when 𝑝 is equal to seven, the point 𝐵 is the same as point 𝐶. This does not make a right-angled triangle, so the only value for 𝑝 we can accept is 𝑝 is equal to four.

Now this is generally the best way to solve a problem like this. However, if we’d sketched this out at the start, we could’ve used some logic in this case. Points 𝐴 and 𝐶 are very easy to sketch. Notice that 𝐵 has a 𝑦-coordinate of three. This means it sits on the same horizontal line as point 𝐶. The only way for the angle at 𝐵 to be a right angle is if it also sits on the same vertical line as point 𝐴. This means it must have the same 𝑥-coordinate as point 𝐴, which is four. Once again, we’ve shown that 𝑝 is equal to four.

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