(i) Draw the graph of x + y = 6 which intersects the X-axis and the Y-

1.	(i) Draw the graph of x + y = 6 which intersects the X-axis and the Y-axis at A and B respectively. Find the length of seg AB. Also find the area of ∆AOB, where O is the point of origin.
Answer» In ∆ AOB, by Pythagoras THEOREM, AB2 = OB2 + OA2 = 62 + 62 = 2 × 36 ∴ AB = 6 √2 OR, A(6, 0) and B(0, 6) D(A,B)= $\sqrt{x2} -x1 {)}^{2} + (y2 - y1 {)}^{2} \\ \\ \\ \sqrt{(0 - 6 {)}^{2} } + (6 - 0 {)}^{2} \\ \\ \\ \sqrt{36 + 36 = \sqrt{72} } = 6 \sqrt{2}$ A(∆ AOB) = 1 / 2 × PRODUCT of sides MAKING right ANGLE = 1 / 2 × 6 × 6 = 18 I *think* it will be helpful to you my bestie(✿^‿^)

(i) Draw the graph of x + y = 6 which intersects the X-axis and the Y-axis at A and B respectively. Find the length of seg AB. Also find the area of ∆AOB, where O is the point of origin.

Answer»

In ∆ AOB, by Pythagoras THEOREM, AB2 = OB2 + OA2 = 62 + 62 = 2 × 36 ∴ AB = 6 √2 OR, A(6, 0) and B(0, 6)

D(A,B)=

$\sqrt{x2} -x1 {)}^{2} + (y2 - y1 {)}^{2} \\ \\ \\ \sqrt{(0 - 6 {)}^{2} } + (6 - 0 {)}^{2} \\ \\ \\ \sqrt{36 + 36 = \sqrt{72} } = 6 \sqrt{2}$

A(∆ AOB) = 1 / 2 × PRODUCT of sides MAKING right ANGLE = 1 / 2 × 6 × 6 = 18

I think it will be helpful to you my bestie(✿^‿^)