Question

ABC is a right angled triangle in which ABC =90 degree and c lies on the x-axis. find the coordinate of c and the area of the triangle

Answer 1

Step-by-step explanation:

Let the point C be (h,K)

Let the point C be (h,K)and A≡(l,0),B≡(0,m)

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)Using distance formula,

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)Using distance formula,a=(h−l)2+K2

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)Using distance formula,a=(h−l)2+K2⇒h−l=a2−K2⟶(3)

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)Using distance formula,a=(h−l)2+K2⇒h−l=a2−K2⟶(3)Similarly

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)Using distance formula,a=(h−l)2+K2⇒h−l=a2−K2⟶(3)SimilarlyK−m=b2−h2⟶(4)

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)Using distance formula,a=(h−l)2+K2⇒h−l=a2−K2⟶(3)SimilarlyK−m=b2−h2⟶(4)Slope of AC=a2−K2K

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)Using distance formula,a=(h−l)2+K2⇒h−l=a2−K2⟶(3)SimilarlyK−m=b2−h2⟶(4)Slope of AC=a2−K2KSlope of BC=hb2−h2

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)Using distance formula,a=(h−l)2+K2⇒h−l=a2−K2⟶(3)SimilarlyK−m=b2−h2⟶(4)Slope of AC=a2−K2KSlope of BC=hb2−h2Since AB is perpendicular to BC

Let the point C be (h,K)and A≡(l,0),B≡(0,m)Slope of AC=h−lk⟶(1)Slope of BC=hK−m⟶(2)Using distance formula,a=(h−l)2+K2⇒h−l=a2−K2⟶(3)SimilarlyK−m=b2−h2⟶(4)Slope of AC=a2−K2KSlope of BC=hb2−h2Since AB is perpendicular to BC∴   a2−K2K.h