Typescript: use either the one or other interface based on property value

Question

I want to have a type that implements either the one or the other properties based on a given other property.

The type should have the following base structure:

interface LinkBase {
  label: string;
  isTextLink: boolean;
}

Depeinding on isTextLink it should have either linkReference or linkText as additional non-optional property:

// if isTextLink is false
export interface LinkWithLinkReference extends LinkBase {
  linkReference: LinkReference;
}

// if isTextLink is true
export interface LinkWithLinkText extends LinkBase {
   linkText: string;
}

I tried to create a union type where either the one or the other interface is allowed:

export type Link= LinkWithLinkReference | LinkWithLinkText;

but it does not take isTextLink value into account and in usage it gives me the following error:

"Property 'linkText' does not exist on type 'Link'.   Property 'linkText' does not exist on type 'LinkWithLinkReference'."

What would be the best way to create and use such type?