Question

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1.If d(P,R) =7; d(P,Q)=10; d(Q,R) = 3 then find if P,Q & R are collinear​

Answer 1

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Answer 2

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Given, Point P(7,7),Q(3,4),R(h,k) are collinear

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4 

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4          = 43=h−3k−4           ...(I)

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4          = 43=h−3k−4           ...(I)It gives, k=43h+7

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4          = 43=h−3k−4           ...(I)It gives, k=43h+7Also, (h−7)2+(k−7)2=100         ...(II)        ∵PR=10

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4          = 43=h−3k−4           ...(I)It gives, k=43h+7Also, (h−7)2+(k−7)2=100         ...(II)        ∵PR=10Now, by substituting the value of k in equation 2 and we got the resultant quadratic equation as follows:

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4          = 43=h−3k−4           ...(I)It gives, k=43h+7Also, (h−7)2+(k−7)2=100         ...(II)        ∵PR=10Now, by substituting the value of k in equation 2 and we got the resultant quadratic equation as follows:h2−14h−15=0

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4          = 43=h−3k−4           ...(I)It gives, k=43h+7Also, (h−7)2+(k−7)2=100         ...(II)        ∵PR=10Now, by substituting the value of k in equation 2 and we got the resultant quadratic equation as follows:h2−14h−15=0Now by solving the above quadratic equation we got, 

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4          = 43=h−3k−4           ...(I)It gives, k=43h+7Also, (h−7)2+(k−7)2=100         ...(II)        ∵PR=10Now, by substituting the value of k in equation 2 and we got the resultant quadratic equation as follows:h2−14h−15=0Now by solving the above quadratic equation we got,         ∴h=15 or h=−1

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4          = 43=h−3k−4           ...(I)It gives, k=43h+7Also, (h−7)2+(k−7)2=100         ...(II)        ∵PR=10Now, by substituting the value of k in equation 2 and we got the resultant quadratic equation as follows:h2−14h−15=0Now by solving the above quadratic equation we got,         ∴h=15 or h=−1            k=13 or k=1

Given, Point P(7,7),Q(3,4),R(h,k) are collinear         ∴3−74−7=h−3k−4          = 43=h−3k−4           ...(I)It gives, k=43h+7Also, (h−7)2+(k−7)2=100         ...(II)        ∵PR=10Now, by substituting the value of k in equation 2 and we got the resultant quadratic equation as follows:h2−14h−15=0Now by solving the above quadratic equation we got,         ∴h=15 or h=−1            k=13 or k=1∴ The point R can be (15,13) & (−1,1)