Step 1: Draw the perpendiculars from coordinates P, Q, and R to X-axis at A, B, and C respectively.
Step 2: Now if we look at the figure carefully, three different trapeziums are formed such as PQAB, PBCR, and QACR in the coordinate plane.
Step 3: So the area of ∆QPR is calculated as
Area of ∆PQR = [Area of trapezium PQAB + Area of trapezium PBCR] - [Area of trapezium QACR] . . . (1)
Step 4: Now calculating areas of all 3 trapeziums.
Since Area of a trapezium = (1 / 2) (sum of the parallel sides) × (distance between sides)
Finding Area of a Trapezium PQAB
⇒ Area of trapezium PQAB = (1 / 2)(QA + PB) × AB
⇒ QA = y2
⇒ PB = y1
⇒ AB = OB – OA = x1 – x2
⇒ Area of trapezium PQAB = (1 / 2)(y1 + y2)(x1 – x2 ) . . . (2)
Finding Area of a Trapezium PBCR
⇒ Area of trapezium PBCR =(1 / 2) (PB + CR) × BC
⇒ PB = y1
⇒ CR = y3
⇒ BC = OC – OB = x3 – x1
⇒ Area of trapezium PBCR =(1 / 2) (y1 + y3 )(x3 – x1) . . . (3)
Finding Area of a Trapezium QACR
⇒ Area of trapezium QACR = (1 / 2) (QA + CR) × AC
⇒ QA = y2
⇒ CR = y3
⇒ AC = OC – OA = x3 – x2
⇒ Area of trapezium QACR =(1 / 2)(y2 + y3 ) (x3 – x2 ). . . (4)
Step 5: Substituting (2), (3) and (4) in (1),
⇒ Area of ∆PQR = (1 / 2)[(y1 + y2)(x1 – x2 ) + (y1 + y3 )(x3 – x1) – (y2 + y3 ) (x3 – x2 )]
⇒ Area of ∆PQR = (1 / 2) |[x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)]|
Therefore, this is the formula to find the area of triangle if coordinates are given.
Now comparing the given coordinates with (x1, y1), (x2, y2), and (x3, y3).
Let, (x1, y1) = (1, 2)
⇒ (x2, y2) = (4, 2)
⇒ (x3, y3) = (3, 5)
Now we have to substitute the values in (1 / 2) |[x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)]|
⇒ (1 / 2) [1 (2 – 5 ) + 4 (5 – 2 ) + 3(2 – 2)]
⇒ (1 / 2) [(- 3) + 12 + 0]
⇒ (1 / 2) [9] = 4.5
Hence the area of the triangle is 4.5 sq units
It is given that the area of the triangle is 1.
Now comparing the given coordinates with(x1, y1), (x2, y2), and (x3, y3).
Let, (x1, y1) = (x1, 1)
⇒ (x2, y2) = (2, 3)
⇒ (x3, y3) = (4, 5)
Now we have to substitute the values in (1 / 2) |[x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)]|
⇒ (1 / 2) [x1(3 - 5 ) + 2(5 - 1 ) + 4(1 - 3)] = 1
⇒ (1 / 2) [x1(- 2) + 8 + -8] = 1
⇒ -x1 = 1
⇒ x1 = -1
Hence, the value of x1 is -1.